Page 2 :
UNIT-3.TRIGONOMETRY, , 3.1, , INTRODUCTION:, , The literal meaning of the word trigonometry is the ‘science of triangle measurement’. The, word ‘trigonometry’ is derived from two greek words ‘trigon’ and ‘metron’ which means, measuring the sides of a triangle. It had its beginning more than two thousand years ago as a tool, for astronomers. The Babylonians, Egyptians, Greeks and the Indians studied trigonometry only, because it helped them in unraveling the mysteries of the universe. In modern times, it has gained, wider meaning and scope. Presently, it is defined as that branch of mathematics which deals with, the measurement of angles, whether of triangle or any other figures., At present trigonometry is used in surveying, astronomy, navigation, physics, engineering, etc., [Trigon → 3 angles/ sides, Metry → Measurement  Trigonometry → Measurement of triangle], 3.2, , RECAPITULATION OF TRIGONOMETRY [Studied in Previous Classes], 1. MEASUREMENT OF ANGLES – RADIAN MEASURES, , Angle: An angle is a figure obtained by rotating a ray about its initial point. If a ray OA is rotated, about its initial point O and reaches to final position OB then the figure obtained is called an angle, AOB and it is denoted by AOB .Here OA is called initial line, OB is called terminal line and O is, called the vertex of an angle. An angle has positive measure if it is measure in anti-clockwise, direction; otherwise it is a negative measure., B, , Terminal line, O, , Angle, Initial line, , A, , An angle can be measured in any one of the following systems:, 1., Sexagesimal system: In this system a right angle is divided into 90 equal parts; each part is, equal to one degree. A degree is divided into 60 equal parts; each part is equal to one minute. A, minute is divided into 60 equal parts; each part is equal to one second., A right angle = 90o, 1/ = 60I, 1’ = 60//,Unit= Degree, 2., Centesimal system: In this system a right angle is divided into 100 equal parts; each part is, equal to one grade. A grade is divided into 100 equal parts; each part is equal to one minute. A, minute is divided into 100 equal parts; each part is equal to one second., A right angle = 100g, 1g = 100/, 1’ = 100//, Unit=Grade, 3., Radian or Circular measure: “A radian is the angle subtended at the centre of the circle, by an arc whose length is equal to the radius of the circle”., A right angle =, , c, 2, , radian, ,  90o = 100g =, , c, 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 2

Page 3 :
Theorem-1: A radian is a constant angle,  radians = 180o, , or, , (, , , radians = 90o), 2, , A radian is less than 60o, 1 radian = 57o 17/ 45//, , or, , Theorem-2: If  is the angle subtended at the centre of the circle by an arc of length S then, S = r, S = Arc length, r = Radius of Circle,  = Angle in radian, Theorem-3: Area of the sector is A =, , 1 2, r, 2, , where A = Area of sector, 1, 2, , Also, A= rs r = Radius of Circle,  = Angle in radian, Radian measure of some common angles, Degree, , 15o 30o, , Radians, , , 12, , , 6, , 45o, , 60o, , , 4, , , 3, , 75o 90o, 5, 12, , 105o, , 120o, , 138o, , 180o, , 165o, , 7, 12, , 2, 3, , 3, 4, , 5, 6, , 11, 12, , , 2, , 180o 270o 360o, , , 3, 2, , 2, , Conversions:, i) To convert degree into radians multiply by, , , 180, , ii) To convert radians into degrees multiply by, Ex:, , 180, , ,  xo = x ,  yc = y , , , 180, 180, , , , radian, degree, , (i) 1200, 241, 2411, = 123o +, , 24 o, 24 o, +, 60 3600, , = 1230 + 0.40 + 0.00670, = 23.40670, = 123.4067 x, , c, 180, , = 2.148c, 2. TRIGNOMETRIC RATIOS, , The trigonometric ratios of a triangle are also called the trigonometric functions. Sine, cosine, and, tangent are 3 important trigonometric functions and are abbreviated as sin, cos and tan. Let us see, how are these ratios or functions, evaluated in case of a right-angled triangle., Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 3

Page 4 :
Consider a right-angled triangle, where the longest side is called the hypotenuse, and the sides, opposite to the hypotenuse are referred to as the adjacent and opposite sides., A, , Hypotenuse= r, , Opposite Side=y, , 90 0, , B, , , , Adjacent Side=x, , C, , The six important trigonometric functions (trigonometric ratios) are calculated using the below, formulas and considering the above figure. It is necessary to get knowledge about the sides of the, right angled triangle because it defines the set of important trigonometric functions., , Functions, , Abbreviation, , Sine Function, Cosine Function, Tangent Function, Cotangent Function, Secant Function, Cosecant Function, , sin, cos, tan, cot, sec, cosec, , Relationship to sides of a, right angled triangle, Opposite side/ Hypotenuse, Adjacent side / Hypotenuse, Opposite side / Adjacent side, Adjacent side / Opposite side, Hypotenuse / Adjacent side, Hypotenuse / Opposite side, , i.e, (i), , Sin =, , Opp y, =, Hyp, r, , (ii), , Cos =, , Adj. x, =, Hyp . r, , (iii), , tan =, , Opp. y, =, Adj. x, , (iv), , Cot =, , Adj. x, =, Opp. y, , (v), , Sec =, , Hyp . r, =, Adj. x, , (vi), , Cosec =, , Hyp . r, =, Opp. y, , Reciprocal relations:, (i), , sin =, , 1, cos ec, , or, , cosec =, , (ii), , cos =, , 1, sec , , or, , sec =, , (iii), , tan =, , 1, cot , , or, , sin , cos , , (ii), , 1, sin , , or, , sin.cosec=1, , 1, cos , , or, , cos.sec=1, , cot =, , 1, tan , , or, , tan.cot=1, , cot =, , cos , sin , , Quotient Relations:, (i), , tan =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 4

Page 5 :
Pythagorean Relations:, Identities:, i) sin2 + cos2 = 1, , ii) 1+tan2 = sec2, , iii) 1+cot2 = cosec2, , Extra Formulae:, sin2 + cos2 = 1, 1+tan2 = sec2, 1+cot2 = cosec2, , , , sin2 = 1 − cos2  sin = 1 − cos 2 , , , , cos2 = 1 − sin2  cos = 1 − sin 2 , , , , tan2 = sec2 − 1  tan = sec 2  − 1, , , , sec2 - tan2 =1  (sec + tan) (sec − tan) = 1, , , , cot2 = cossec2 − 1  cot = cos ec 2  − 1, , , cosec2 - cot2 = 1  (cosec + cot) (cosec − cot) = 1, Complementary and supplementary angles:, i) Two angles are said to be complementary when their sum is equal to right angle (i.e. 90o), ii) sin and cos, tan and cot, sec and cosec each ratio of a pair being called co-ratio of the other ratio., Any ratio of an acute angle is equal to corresponding co-ratio of the complementary angle., Ex:, sin30o = cos60o, tan75o = cot15o’, sec63o = cosec27o, etc., iii) Two angle are said to be supplementary angles when their sum is equal to 180o, iv) If two angles differ by an integral multiple of 360o then two angles are said to be conterminal, angles, 3. TRIGONOMETRIC RATIOS OF STANDARD ANGLES:, Standard angle table:, , , , Degree, Radian, , 00, 0, , 45 0, , 4, 1, 2, , 3, 2, 1, 3, , 1, 2, , 3, 2, 1, 2, , 1, , 3, , , , 1, 3, 2, , 0, , sin , , 0, , cos , , 1, , tan , , 0, , cot , , , , 3, , 1, , sec , , 1, , 2, , cosec , , , , 2, 3, 2, , 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , 60 0, , 3, , 90 0, , 2, 1, , 30 0, , 6, 1, 2, , 2, 3, , 0, , , , 1, , Page 5

Page 6 :
3.3 ALLIED ANGLES, , 3.3, , Allied Angles:, , , , 2, , , , 3.3.1 Definition of Allied Angles: Angles of the form  n.    are called Allied angles. In, , , , , particular, angles of the form (90o±), (180º±), (270º±), (360º±) are called “Allied angles”.,  c, 90 0 ,  2, , , 3.3.2 ASTC-Rule:, S, 90 + , 180 − , , 90º, , , , , , , A, 9(− , 90, 360 + , , sin & cosec are +ve, , All are +ve, , 180º(c), , 180 + , , 360 − , , 270 − , , 270 + , cos & sec are +ve, , tan & cot are +ve, 270º, , T, ,  3c, ,  2, , , •, , 00 ,, 360º(2c), , , , , , , C, , Sign of trigonometric ratios:, Depending on the value of angle  trigonometric ratios takes different signs in different, quadrants as x and y is positive or negative corresponding to different quadrants but, hypotenuse r is always positive. The sign all trigonometric ratios in all four quadrants are, given by, , , Quadrant, , I (A), , II (S), , III (T), , IV (C), , sin & cosec, , + ve, , + ve, , − ve, , − ve, , cos & sec, , + ve, , − ve, , − ve, , + ve, , tan & cot, , + ve, , − ve, , + ve, , − ve, , i.e, 1., , In the first quadrant all the trigonometric ratios are positive in sign, and angles lies in it are 90 −  & 360 + , , 2., , In the second quadrant sine and its reciprocal ratio co-sine are positive in sign,, remaining ratios are negative in sign and the angles lies in it are 90 +  & 180 − , , 3., , In the third quadrant tangent and its reciprocal ratio co-tangent are positive in sign,, remaining ratios are negative in sign and the angles lies in it are 180 +  & 270 − , , 4., , In the fourth quadrant cosine and its reciprocal ratio secant are positive in sign,, remaining ratios are negative in sign and the angles lies in it are 270 +  & 360 − , , 5., , It is remembered as ASTC-rule., , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 6

Page 7 :
3.3.3 Rules of Allied angles:, 1., , In case of allied angles like 90 +  ,90 −  ,270 +  & 270 −  , the trigonometric functions, changes to its co-functions and sign is written by ASTC-Rule:, Trigonometric functions, Sine, Co-sine, tangent, Co-tangent, Secant, Co-secant, , Co-functions, Co-sine, Sine, Co-tangent, tangent, Co-secant, Secant, , 2., , In case of allied angles like 180 +  ,180 −  ,360 +  & 360 −  , the trigonometric functions, , 3., , remains as it is and sign is written by ASTC-Rule., The trigonometric functions of −  is same as trigonometric functions of 360 − , sin (−  ) = sin (360 −  ) = − sin , , cos(−  ) = cos(360 −  ) = cos , , tan (−  ) = tan (360 −  ) = − tan , , cot(−  ) = cot(360 −  ) = − cot , , sec(−  ) = sec(360 −  ) = sec , , cos ec(−  ) = cos ec(360 −  ) = − cos ec, ➢ Trigonometric ratios of 90 −  & 360 +  [I-Quadrant] in terms of an angle  using above, rules:, 1. sin( 90 −  ) = cos , , 2. cos(90 −  ) = sin , , 5. sec(90 −  ) = cos ec, , 6. cos ec(90 −  ) = sec , , 1. sin( 360 +  ) = sin , , 2. cos(360 +  ) = cos , , 5. sec(360 +  ) = sec , , 6. cos ec(360 +  ) = cos ec, , 3. tan (90 −  ) = cot , , 3. tan (360 +  ) = tan , , 4. cot (90 −  ) = tan , , 4. cot (360 +  ) = cot , , ➢ Trigonometric ratios of 90 +  & 180 −  [II-Quadrant] in terms of an angle  using above, rules:, , 1. sin( 90 +  ) = cos , , 2. cos(90 +  ) = − sin , , 5. sec(90 +  ) = − cos ec, , 6. cos ec(90 +  ) = sec , , 1. sin( 180 −  ) = sin , , 2. cos(180 −  ) = − cos , , 5. sec(180 −  ) = − sec , , 6. cos ec(180 −  ) = cos ec, , 3. tan (90 +  ) = − cot , , 3. tan (180 −  ) = − tan , , 4. cot (90 +  ) = − tan , , 4. cot (180 −  ) = − cot , , ➢ Trigonometric ratios of 180 +  & 270 −  [III-Quadrant] in terms of an angle  using above, rules:, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 7

Page 8 :
1. sin( 180 +  ) = − sin , , 2. cos(180 +  ) = − cos , , 5. sec(180 +  ) = − sec , , 6. cos ec(180 +  ) = − cos ec, , 1. sin( 270 −  ) = − cos , , 2. cos(270 −  ) = − sin , , 5. sec(270 −  ) = − cos ec, , 6. cos ec(270 −  ) = − sec , , 3. tan (180 +  ) = tan , , 4. cot (180 +  ) = cot , , 3. tan (270 −  ) = cot , , 4. cot (270 −  ) = tan , , ➢ Trigonometric ratios of 270 +  & 360 −  [IV-Quadrant] in terms of an angle  using above, rules:, , 1. sin( 270 +  ) = − cos , , 2. cos(270 +  ) = sin , , 5. sec(270 +  ) = cos ec, , 6. cos ec(270 +  ) = − sec , , 1. sin( 360 −  ) = − sin , , 2. cos(360 −  ) = cos , , 5. sec(360 −  ) = sec , , 6. cos ec(360 −  ) = − cos ec, , 3. tan (270 +  ) = − cot , , 4. cot (270 +  ) = − tan , , 3. tan (360 −  ) = − tan , , 4. cot (360 −  ) = − cot , , Table of trigonometric ratios of allies angles in terms of acute angles:, Quadrant, , sin, , cos, , tan, , cot, , sec, , cosec, , 90 0 − , , cos , , sin , , cot , , tan , , cosec , , sec , , 90 0 + , , cos , , -sin , , -cot , , -tan , , -cosec , , sec , , 180 0 − , , sin , , -cos , , -tan , , -cot , , -sec , , cosec , , 180 0 + , , -sin , , -cos , , tan , , cot , , -sec , , -cosec , , 270 0 − , , -cos , , -sin , , cot , , tan , , -cosec , , -sec , , 2700 + , , -cos , , sin , , -cot , , -tan , , cosec , , -sec , , 360 0 − , , -sin , , cos , , -tan , , -cot , , sec , , -cosec , , 360 0 + , , sin , , cos , , tan , , cot , , sec , , cosec , , −, , -sin , , cos , , -tan , , -cot , , sec , , -cosec , , 3.3.5 Problems on finding the numerical values of trigonometric functions:, 1., Find the value of trigonometric functions of an angle 1200, Solution:, , (, , ), , sin 1200 = sin 180 − 600 = sin 600 =, , 3, 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 8

Page 9 :
1, (, ), 2, tan 120 = tan (180 − 60 ) = − tan 60 = − 3, 1, cot 120 = cot (180 − 60 ) = − cot 60 = −, 3, sec 120 = sec(180 − 60 ) = − sec 60 = −2, cos ec120 = cos ec(180 − 60 ) = cos ec60 =, cos 120 0 = cos 180 − 60 0 = − cos 60 0 = −, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 2, 3, , 2., Find the value of trigonometric functions of an angle 1350, Solution:, 1, sin 1350 = sin (180 − 450 ) = sin 450 =, 2, 1, cos 1350 = cos(180 − 450 ) = − cos 450 = −, 2, tan 1350 = tan (180 − 450 ) = − tan 450 = −1, , (, ), sec 135 = sec(180 − 45 ) = − sec 45 = − 2, cos ec135 = cos ec(180 − 45 ) = cos ec 45 =, cot 1350 = cot 180 − 450 = − cot 450 = −1, 0, , 0, , 0, , 0, , 0, , 0, , 2, , 3., Find the value of trigonometric functions of an angle 1500, Solution:, 1, sin 150 0 = sin 180 − 30 0 = sin 30 0 =, 2, 3, cos 150 0 = cos 180 − 30 0 = − cos 30 0 = −, 2, 1, tan 150 0 = tan 180 − 30 0 = − tan 30 0 = −, 3, , (, , ), , (, , ), , (, , ), , (, ), 2, sec 150 = sec(180 − 30 ) = − sec 30 = −, 3, cos ec150 = cos ec(180 − 30 ) = cos ec30 = 2, cot 150 0 = cot 180 − 30 0 = − cot 30 0 = − 3, 0, , 0, , 0, , 0, , 0, , 0, , 4., Find the value of trigonometric functions of an angle 1800, Solution:, sin 180 0 = sin 180 − 0 0 = sin 0 0 = 0, , (, ), cos 180 = cos(180 − 0 ) = − cos 0 = −1, tan 180 = tan (180 − 0 ) = − tan 0 = 0, cot 180 = cot (180 − 0 ) = − cot 0 = −, sec 180 = sec(180 − 0 ) = − sec 0 = −1, cos ec180 = cos ec(180 − 0 ) = cos ec0 = , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 5, Find the value of trigonometric functions of an angle 210 0, Solution:, , (, , ), , sin 2100 = sin 270 − 600 = − sin 600 = −, , 3, 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 9

Page 10 :
1, (, ), 2, tan 210 = tan (270 − 60 ) = tan 60 = 3, 1, cot 210 = cot (270 − 60 ) = cot 60 =, 3, sec 210 = sec(270 − 60 ) = − sec 60 = −2, cos ec 210 = cos ec(270 − 60 ) = − cos ec60, cos 210 0 = cos 270 − 60 0 = − cos 60 0 = −, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , =−, , 2, 3, , 6., Find the value of trigonometric functions of an angle 225 0, Solution:, 1, sin 2250 = sin 180 + 450 = − sin 450 = −, 2, 1, cos 2250 = cos 180 + 450 = − cos 450 = −, 2, 0, 0, 0, tan 225 = tan 180 + 45 = tan 45 = 1, , (, , ), , (, ), (, ), cot 225 = cot (180 + 45 ) = cot 45 = 1, sec 225 = sec(180 + 45 ) = − sec 45 = − 2, cos ec 225 = cos ec(180 + 45 ) = − cos ec 45, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , =− 2, , 7., Find the value of trigonometric functions of an angle 270 0, Solution:, , (, ), cos 270 = cos(270 − 0 ) = − sin 0 = 0, tan 270 = tan (270 − 0 ) = cot 0 = , cot 270 = cot (270 − 0 ) = tan 0 = 0, sec 270 = sec(270 − 0 ) = − cos ec0 = −, cos ec 270 = cos ec(270 − 0 ) = − sec 0 = −1, , sin 270 0 = sin 270 − 0 0 = − cos 0 0 = −1, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 8., Find the value of trigonometric functions of an angle 240 0, Solution:, , (, , ), , 3, 2, 1, cos 240 0 = cos 270 − 30 0 = − sin 30 0 = −, 2, 0, 0, 0, tan 240 = tan 270 − 30 = cot 30 = 3, , sin 240 0 = sin 270 − 30 0 = − cos 30 0 = −, , (, ), (, ), 1, cot 240 = cot (270 − 30 ) = tan 30 =, 3, sec 240 = sec(270 − 30 ) = − cos ec30 = −2, cos ec 240 = cos ec(270 − 30 ) = − sec 30 = −, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 2, 3, , 9., Find the value of trigonometric functions of an angle 300 0, Solution:, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 10

Page 11 :
(, , ), , sin 3000 = sin 360 − 600 = − sin 600 = −, , 3, 2, , 1, (, ), 2, tan 300 = tan (360 − 60 ) = − tan 60 = − 3, 1, cot 300 = cot (360 − 60 ) = − cot 60 = −, 3, sec 300 = sec(360 − 60 ) = sec 60 = 2, cos ec300 = cos ec(360 − 60 ) = − cos ec60 = −, cos 3000 = cos 360 − 600 = cos 600 =, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 2, 3, , 0, , 10., Find the value of trigonometric functions of an angle − 3150, Solution:, , (, , ), , (, , ), , sin − 3150 = − sin 3150 = − sin 360 − 450 = −  − sin 450 =, , 1, 2, , 1, (, ), (, ), 2, tan (− 315 ) = − tan 315 = − tan (360 − 45 ) = −  − tan 45 = 1, cot (− 315 ) = − cot 315 = − cot (360 − 45 ) = −  − cot 45 = 1, sec (− 315 ) = sec 315 = sec (360 − 45 ) = sec 45 = 2, cos ec(− 315 ) = − cos ec315 = − cos ec (360 − 45 ) = −  − cos ec 45, cos − 3150 = cos 3150 = cos 360 − 450 = cos 450 =, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , = 2, , 11., Find the value of trigonometric functions of an angle 360 0, Solution:, , (, ), cos 360 = cos(360 − 0 ) = cos 0 = 1, tan 360 = tan (360 − 0 ) = − tan 0 = 0, cot 360 = cot (360 − 0 ) = − cot 0 = −, sec 360 = sec(360 − 0 ) = sec 0 = 1, cos ec360 = cos ec(360 − 0 ) = − cos ec0 = −, Find the values of i.Sin (− 150 ) ii. tan (− 210 ) and iii . cos(− 330 ), i.Sin (− 150 ), sin 360 0 = sin 360 − 0 0 = − sin 0 0 = 0, 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 12., , 0, , 0, , 0, , 0, , 0, , 0, , Solution :, , (, , ), , (, , ), , sin − 1500 = − sin 1500 = − sin 180 − 300 = − sin 30 0 = −, , (, , ii. tan − 2100, , (, , ), , ), , (, , 1, 2, , ), , tan − 210 0 = − tan 210 0 = − tan 180 + 30 0 = − tan 30 0 = −, , (, , iii . cos − 330 0, , (, , ), , ), , (, , ), , Sin (−  ) = − Sin, , tan (−  ) = − tan , , 1, 3, , 3, 2, and iii . sin − 6900, , cos(−  ) = cos , , cos − 330 0 = cos 330 0 = cos 360 − 30 0 = cos 30 0 =, , 13., , (, , ), , (, , Find the values of i.Co sec − 8400 ii. cot − 3900, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , (, , ), Page 11

Page 12 :
(, , i.Co sec − 8400, Solutions, , (, , ), , ), , (, , ), , Co sec − 8400 = − cos ec8400 = − cos ec 2  360 + 1200 = − cos ec1200, , (, , ), , = − cos ec 180 − 60 0 = − cos ec60 0 = −, , (, , ii. cot − 390 0, Solution :, , (, , ), , ), , (, , 2, 3, , ), , cot − 390 0 = − cot 390 0 = − cot 360 + 30 0 = − cot 30 0 = − 3, , (, , iii . sin − 690, , 0, , ), , Solution :, , (, , ), , (, , ), , sin − 690 0 = − sin 690 0 = − sin 2  360 − 30 0 = −  − sin 30 0 =, , 14., , (, , ), , (, , ), , (, , Find the values of i. sec 11100 ii. cos 17700 and iii . sin 15000, , (, , i. sec 1110 0, , ), , 1, 2, , ), , Solution :, , (, , ), , (, , ), , 2, 3, , (, , ), , 3, 2, , sec 1110 0 = Sec 3  360 + 30 0 = sec 30 0 =, , (, , ii. cos 1770 0, , ), , Solution :, , (, , ), , cos 1770 0 = cos 5  360 − 30 0 = cos 30 0 =, , (, , iii . sin 1500 0, , ), , Solution :, , (, , ), , (, , 3, ), 2, Find the values of i. tan (1560 ) ii. cos ec(− 1110 ) and iii . sec(− 855 ), i. tan (1560 ), sin 1500 0 = sin 4  360 + 60 0 = sin 60 0 =, , 15., , 0, , 0, , 0, , 0, , Solution :, , (, , ), , (, , ), , tan 1560 0 = tan (4  360 + 120) = tan 120 0 = tan 180 − 60 0 = − tan 60 0 = − 3, , (, , ii. cos ec − 1110 0, , ), , Solution :, , (, ), iii . sec(− 855 ), , (, , ), , cos ec − 1110 0 = − cos ec1110 0 = − cos ec 3  360 + 30 0 = − cos ec30 0 = −2, 0, , Solution :, , (, , ), , (, , ), , (, , ), , sec − 8550 = sec 8550 = sec 2  360 + 1350 = sec 1350 = sec 180 − 450 = − sec 450 = − 2, , 16., ,  25, Find the values of i. tan ,  18, , , 5 ,  5 ,  ii. cos ec −,  and iii . sec −, , ,  3 ,  4 , , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 12

Page 13 :
Solution,  25 , i. tan , ,  3 ,  25, tan ,  3, ,  25 1800 , , ,  = tan 15000, =, tan, , 3, , , , , (, , ), , = tan 4  360 + 60 0 = tan 60 0 = 3,  5 , ii. cos ec −, ,  3 ,  5  180 0 ,  5 ,  = − cos ec300 0, cos ec −,  = − cos ec, 3 ,  3 , , , (, , ), , = − cos ec 360 − 60 0 = −  − cos ec60 0 =, , 2, 3, ,  5 , iii . sec −, ,  4 ,  5  180 0 ,  5 ,  = sec 2250, sec −,  = sec, 4 ,  4 , , , (, , ), , = sec 180 + 450 = − sec 450 = − 2, , 3.3.6 Problems on finding the values expression and unknown term involving, trigonometric functions:, 1., , Find the value of sin 510 0 cos 570 0 + cos 390 0 sin 330 0, Solution:, , sin 5100 cos 5700 + cos 3900 sin 3300, , (, , ) (, , ), , (, , ) (, , = sin 360 + 1500 cos 360 + 2100 + cos 360 + 300 sin 360 − 300, , ), , = sin 150 0 cos 210 0 + cos 30 0  − sin 30 0, , (, , ) (, , ), , = sin 180 − 30 0 cos 180 + 30 0 − cos 30 0 sin 30 0, = sin 30  − cos 30 − cos 30 sin 30 0, 0, , 0, , 0, , 1, 3, 3 1, =− , −, , 2 2, 2 2, 3, 3 − 3− 3, =−, −, =, 4, 4, 4, −2 3, 3, =, =−, 4, 2, 2., , Find the value of sin 510 0 cos 570 0 − cos 390 0 sin 330 0, Solution:, , sin 5100 cos 5700 − cos 3900 sin 3300, , (, , ) (, , ), , (, , ) (, , = sin 360 + 1500 cos 360 + 2100 − cos 360 + 300 sin 360 − 300, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , ), Page 13

Page 14 :
= sin 150 0 cos 210 0 − cos 30 0  − sin 30 0, , (, , ) (, , ), , = sin 180 − 30 0 cos 180 + 30 0 + cos 30 0 sin 30 0, = sin 30  − cos 30 + cos 30 sin 30 0, 0, , 0, , 0, , 1, 3, 3 1, =− , +, , 2 2, 2 2, , =−, , 3, 3, +, 4, 4, , =0, 3., , (, , ), , Simplify without using table/calculator: sin 7800 cos 3300 + cos − 1200 sin 3900, Solution:, , (, , ), , sin 780 0 cos 330 0 + cos − 120 0 sin 390 0, , (, , ) (, , ), , (, , ) (, , = sin 2  360 + 60 0 cos 360 − 30 0 + cos 180 − 60 0 sin 360 + 30 0, , ), , = sin 60 0 cos 30 0 +  − cos 60 0 sin 30 0, 3, 3 1 1, , − , 2, 2 2 2, 3 1 3 −1, = − =, 4 4, 4, 2 1, = =, 4 2, , =, , 4., , (, , ), , Simplify without using table/calculator: sin 7800 cos 300 + cos 1200 sin 3900, Solution:, , (, , ), , sin 7800 cos 300 + cos 1200 sin 3900, , (, , ), , (, , ) (, , = sin 2  360 + 600 cos 300 + cos 180 − 600 sin 360 + 300, , ), , = sin 600 cos 300 +  − cos 600 sin 300, , 3, 3 1 1, , − , 2, 2 2 2, 3 1 3 −1, = − =, 4 4, 4, 2 1, = =, 4 2, , =, , 5., , (, , ), , Show that sin 6000 cos 3300 + cos 1200 sin 1500 = −1, Solution:, , (, , ), , L.H .S = sin 6000 cos 3300 + cos 1200 sin 1500, , (, , ) (, , ), , (, , ) (, , = sin 360 + 2400 cos 360 − 300 + cos 180 − 600 sin 180 − 300, , ), , = sin 2400 cos 300 +  − cos 600 sin 300, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 14

Page 15 :
(, , ), , = sin 180 + 60 0 cos 30 0 − cos 60 0 sin 30 0, = − sin 60 0 cos 30 0 − cos 60 0 sin 30 0, 3 3 11, −, 2 2 22, 3 1 − 3 −1, 4, =− − =, =−, 4 4, 4, 4, = −1, , =−, , = R.H .S, 6., , (, , ), , Show that sin 4200 cos 3900 + cos 3000 sin − 3300 = 1 without using table/calculator., Solution:, , (, , L.H .S = sin 420 0 cos 390 0 + cos 300 0 sin − 330 0, , (, , ) (, , ), , (, , ), , ), , (, , = sin 360 + 60 0 cos 360 + 30 0 + cos 360 − 60 0  − sin 360 − 30 0, , ), , = sin 60 0 cos 30 0 − cos 60 0  − sin 30 0, 3, 3 1 1, , + , 2, 2 2 2, 3 1 3 +1 4, = + =, =, 4 4, 4, 4, =1, =, , 7., , Show that tan 2250 cot 4050 + tan 7650 cot 6750 + cos ec1350 sec 3150 = 2, , Solution:, tan 2250 = tan(180 + 45) = tan 450 = 1, cot 4050 = cot(360 + 450 ) = cot 450 = 1, tan 7650 = tan( 2  360 + 450 ) = tan 450 = 1, cot 6750 = cot( 2  360 − 450 ) = − cot 450 = −1, , cos ec1350 = cos ec(180 − 450 ) = cos ec 450 = 2, sec 3150 = sec(360 − 450 ) = sec 450 = 2, L.H .S = tan 2250 cot 4050 + tan 7650 cot 6750 + cos ec1350 sec 3150, = 1  1 + 1  −1 + 2  2, = 1−1+ 2, = 2 = R.H .S, tan 2 1200 cot 2 2400 − cot 2 2100 tan 2 2400, 8., Find the value of, 3 tan 450 sin 2700 − sec 3600 cos 2700, Solution:, tan 1200 = tan 180 − 600 = − tan 600 = − 3, , cot 2400, , (, ), = cot (180 + 60 ) = cot 60, 0, , 0, , =, , 1, 3, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 15

Page 16 :
(, ), = tan (180 + 60 ) = tan 60, , cot 210 0 = cot 180 + 30 0 = cot 30 0 = 3, tan 240 0, , 0, , 0, , = 3, , tan 450 = 1, , (, ), = sec(360 − 0 ) = sec 0 = 1, = cos(180 + 90 ) = − cos 90, , sin 270 0 = sin 180 + 90 0 = − sin 90 0 = −1, sec 360 0, cos 270 0, , 0, , 0, , 0, , 0, , = −0 = 0, , tan 120 cot 240 0 − cot 2 210 0 tan 2 240 0, Given exp ression =, 3 tan 450 sin 270 0 − sec 360 0 cos 270 0, 2, , 0, , 2, , (− 3 )  13  − ( 3 ) ( 3 ), 2, , 2, , 2, , 2, , , , 3(1)(− 1) − 1(0 ), −8 8, =, =, −3 3, sin 1350 − cos 120 0, 9., Show that, = 3 + 2 2., sin 1350 + cos 480 0, Solution:, =, , (, , ), , (, , ), , 1, 2, , sin 1350 = sin 180 − 450 = sin 450 =, , cos 120 0 = cos 180 − 60 0 = − cos 60 0 = −, , (, , 1, 3 − 3 3, 1− 9, 3, =, =, −3−0, −3, , 1, 2, , ), , (, , ), , cos 480 0 = cos 360 + 120 0 = cos 120 0 = cos 180 − 60 0 = − cos 60 0, sin 1350 − cos 120 0, sin 1350 + cos 480 0, 1, 1, 2+ 2, +, 2+ 2, 2, = 2, = 2 2 =, 1, 1, 2− 2, 2− 2, −, 2 2, 2 2, , =−, , 1, 2, , L.H .S =, , (, , ), , 2, , 2+ 2 2+ 2, 2+ 2, =, , =, 2 − 2 2 + 2 (2 )2 − 2, , (, , 6+ 4 2 2 3+ 2 2, =, 4−2, 2, = 3 + 2 2 = R.H .S, , =, , 10., , If, , ( ), , 2, , =, , 4 + 2 + 2 2 2, 4−2, , ), , x sin 2 3000 cos 2 3150, = cos ec 2 2250 find x., 2, 0, 2, 0, tan 315 cot 405, , Solution:, , (, , ), , sin 3000 = sin 360 − 600 = − sin 600 = −, , (, ), = tan (360 − 45 ) = − tan 45, , 3, 2, , 1, 2, = −1, , cos 3150 = cos 360 − 450 = cos 450 =, tan 3150, , 0, , 0, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 16

Page 17 :
(, , ), , cot 4050 = cot 360 + 450 = cot 450 = 1, , (, , ), , cos ec2250 = cos ec 180 + 450 = − cos ec450 = − 2, x sin 2 300 0 cos 2 3150, = cos ec 2 2250, 2, 0, 2, 0, tan 315 cot 405, 2, , , 3  1 , 3 1,  , x −, x, , ,   2 , 2, 2, , , 4, 2 =2, = − 2 , 1, (− 1)2 (1)2, 3x, = 2  3 x = 16, 8, 16, x=, 3, x cos 2 3000 cos ec 2 240 0, 11., Find the value of x if, = tan 2 3150 cot 2 300 0, sin 2 2250 sec 2 240 0, Solution:, 1, cos 300 0 = cos 360 − 60 0 = cos 60 0 =, 2, 2, cos ec 240 0 = cos ec 180 + 60 0 = − cos ec60 0 = −, 3, 1, sin 2250 = sin 180 + 450 = − sin 450 = −, 2, 0, 0, 0, sec 240 = sec 180 + 60 = − sec 60 = −2, 2, , (, , ), , (, , ), , (, , tan 3150, cot 300 0, , ), , (, ), (, ), = tan (360 − 45 ) = − tan 45, = cot (360 − 60 ) = − cot 60, 0, , 0, , = −1, , 0, , 0, , =−, , 1, 3, , x cos 2 300 0 cos ec 2 240 0, = tan 2 3150 cot 2 300 0, 2, 0, 2, 0, sin 225 sec 240, 2, , 12., , 2, , 1  2 , x   −, , 2, 1 , 3, 2 , 2, = (− 1)  −, , 2, 3, ,  1 , 2, −,  (− 2), 2, , 1 4, x , 4 3 = 1 1, 1, 3, 4, 2, x/3 1, =, 2, 3, x 1, =, 6 3, 1, x =  6 = 2., 3, ,   3 ,  , Simplify sin cos −  − cos sin  −, , 3, 4  4 ,  3, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 17

Page 18 :
, , 180 0, 3, = sin 60 0 =, 3, 3, 2,  180 0 , 1,  ,  = cos 60 0 =, cos −  = cos −, 3 , 2,  3, , , sin, , cos, , , 4, , = sin, , = cos, , 180 0, 1, = cos 450 =, 4, 2, ,  3  180 0 , 1,  3 , ,  = − sin 1350 = − sin 180 − 450 = − sin 450 = −, sin  −, =, −, sin, , 4 , 2,  4 , , Consider given expression, ,   3 ,  , sin cos −  − cos sin  −, , 3, 4  4 ,  3, , (, , =, , ), , 3 1, 1, 1,  −, −, 2 2, 2, 2, , 3 1, 3+2, + =, 4 2, 4,  ,  3 ,  5 ,  7 , 13., Show that sin 2   + sin 2   + sin 2   + sin 2 , =2, 4,  4 ,  4 ,  4 , Solution:, 1,  , sin   = sin 450 =, 2, 4, 1,  3 , sin   = sin 1350 = sin (180 − 450 ) = sin 450 =, 2,  4 , =, ,  5 , sin   = sin 2250 = sin 180 + 450 = − sin 450 = −,  4 ,  7 , 0, 0, 0, sin ,  = sin 315 = sin 360 − 45 = − sin 45 = −,  4 , , (, , ), , (, , ), , 1, 2, 1, 2, ,  ,  3 ,  5 ,  7 , L.H .S = sin 2   + sin 2   + sin 2   + sin 2 , , 4,  4 ,  4 ,  4 , 2, , 2, , 2, , 2, ,  1   1   1   1 , =,  +,  + −,  + −, , 2 , 2,  2  2 , 1 1 1 1, = + + +, 2 2 2 2, 1+1+1+1 4, =, = = 2 = R.H .S, 2, 2, ,  ,  3 ,  5 ,  7 , 14., Show that cos 2   + cos 2   + cos 2   + ocs 2 , =2, 4,  4 ,  4 ,  4 , Solution:, 1,  , cos  = cos 450 =, 2, 4, 1,  3 , cos  = cos 1350 = cos(180 − 450 ) = − cos 450 = −, 2,  4 , 1,  5 , cos  = cos 2250 = cos 180 + 450 = − cos 450 = −, 2,  4 , , (, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 18

Page 19 :
1,  7 , 0, 0, 0, cos,  = cos 315 = cos 360 − 45 = cos 45 =, 2,  4 ,  ,  3 ,  5 ,  7 , L.H .S = cos 2   + cos 2   + cos 2   + ocs 2 , , 4,  4 ,  4 ,  4 , , (, , 2, , 2, , ), , 2, , 2, ,  1   1   1   1 , =,  + −,  + −,  +, , 2 , 2  2,  2 , 1 1 1 1, = + + +, 2 2 2 2, 1+1+1+1 4, =, = = 2 = R.H .S, 2, 2,  ,  3 ,  5 ,  7 , 15., Show that tan 2   + tan 2   + tan 2   + tan 2 , =4, 4,  4 ,  4 ,  4 , Solution:,  , tan   = tan 450 = 1, 4,  3 , tan   = tan 1350 = tan (180 − 450 ) = − tan 450 = −1,  4 ,  5 , tan   = tan 2250 = tan (180 + 450 ) = tan 450 = 1,  4 ,  7 , 0, 0, 0, tan ,  = tan 315 = tan (360 − 45 ) = − tan 45 = −1, 4, , ,  ,  3 ,  5 ,  7 , L.H .S = tan 2   + tan 2   + tan 2   + tan 2 , , 4,  4 ,  4 ,  4 , = (1) + (− 1) + (1) + (− 1), 2, , 2, , 2, , 2, , = 1 + 1 + 1 + 1 = 4 = R.H .S,  ,  3 ,  5 ,  7 , Show that cot 2   + cot 2   + cot 2   + cot 2 , =4, 4,  4 ,  4 ,  4 , Solution:,  , cot   = cot 450 = 1, 4,  3 , cot   = cot 1350 = cot 180 − 450 = − cot 450 = −1,  4 , , (, , ), ,  5 , cot   = cot 2250 = cot 180 + 450 = cot 450 = 1,  4 ,  7 , 0, 0, 0, cot ,  = cot 315 = cot 360 − 45 = − cot 45 = −1, 4, , , , (, , ), , (, , ), ,  ,  3 ,  5 ,  7 , L.H .S = cot 2   + cot 2   + cot 2   + cot 2 , , 4,  4 ,  4 ,  4 , = (1) + (− 1) + (1) + (− 1), 2, , 2, , 2, , 2, , = 1 + 1 + 1 + 1 = 4 = R.H .S, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 19

Page 20 :
3.3.7 Problems on simplification using allied angles, 1., , Simplify, , (, (, , ) (, ) (, , ) (, ) (, , ), ), , sin 2700 − A tan 1800 − A sec 3600 + A, cos 1800 − A cos ec 900 − A tan 1800 + A, , Solution:, Consider, sin 270 0 − A tan 180 0 − A sec 360 0 + A, cos 180 0 − A cos ec 90 0 − A tan 180 0 + A, − cos A  − tan A  sceA, =, − cos A  sec A  tan A, 1, =, = −1, −1, cos 900 + , sin (−  ), cot 2700 + , 2., Simplify, +, −, sin 900 − , sin 3600 − , tan 1800 − , Solution:, Consider give expression, cos 90 0 + , sin (−  ), cot 270 0 + , +, −, sin 90 0 − , sin 360 0 − , tan 180 0 − , − sin  − sin  − tan , =, +, −, cos , − sin  − tan , = − tan  + 1 − 1, = − tan , cos(−  ), tan 900 − , cos 900 + , 3., Simplify, −, +, sin 2700 +  cot 1800 −  cos 2700 − , Solution:, Consider the given expression, cos(−  ), tan 90 0 − , cos 90 0 + , −, +, sin 270 0 +  cot 180 0 − , cos 270 0 − , cos , cot , − sin , =, −, +, − cos  − cot  − sin , 1, 1 −1, =, −, +, −1 −1 −1, = −1 + 1 + 1 = 1, cos ec 1800 − A cos(− A), 4., Show that, = cot 2 A, 0, 0, sec 180 + A cos 90 + A, Solution:, cos ec(180 0 − A)cos(− A), L.H .S =, sec(180 0 + A)cos(90 0 + A), cos ecA  cos A, 1, cos A, =, =,  cos A , − sec A  − sin A sin A, sin A, 2, cos A, =, = cot 2 A, 2, sin A, sin 1800 + A cos 3600 − A tan 1800 + A, 5., Show that, = −1, cos 2700 + A sin 900 + A cot 2700 − A, Solution:, , (, (, , ) (, ) (, , (, (, , ), ), , (, (, , (, , ), ), , ), , (, , (, (, , (, , (, (, , (, (, , (, (, , ), , ) (, , ) (, ) (, , ), ), , (, (, , ), , ), , ), ), , ), , (, , (, , (, , ) (, ) (, , ), ), , ), ), , ), ), , (, (, , (, (, , ), ), , ), ), , ), , ) (, ) (, ,  cos ecA =, , 1, sin A, , 1, = cos A, sec A, , ), ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 20

Page 21 :
(, (, , ) (, ) (, , ) (, ) (, , ), ), , sin 180 0 + A cos 360 0 − A tan 180 0 + A, cos 270 0 + A sin 90 0 + A cot 270 0 − A, − sin A  cos A  tan A, =, sin A  cos A  tan A, = −1 = R.H .s, , , tan  + A , sin (2 − A), 2,  + cos ec(− A), 6., Simplify, −, sin ( − A) cot (2 + A), , , sec + A , 2, , Solution:, Consider given expression, , , tan  + A , sin (2 − A), 2,  + cos ec(− A), −, sin ( − A) cot (2 + A), , , sec + A , 2, , − sin A − cot A − cos ecA, =, −, +, sin A, cot A, − cos ecA, = −1 − (− 1) + 1 = −1 + 1 + 1 = 1, L.H .S =, , , , sin ( − A) cot − A  cos(2 − A), 2, , 7., Show that, = sin A, , , tan ( + A) tan  + A  sin (− A), 2, , Solution:, , , sin ( − A) cot  − A  cos(2 − A), 2, , L.H .S =, , , tan ( + A) tan  + A  sin (− A), 2, , sin A  tan A  cos A, =, tan A  − cot A  − sin A, cos A, sin A, 1, sin A, W .K .T, = tan A and tan A =, =, = cos A  tan A = cos A , cot, A, cos, A, cot A, cos A, = sin A = R.H .S,  5, , cos, − ,  2,  + tan (−  ) = sec 2 , 8., Show that, sin (4 +  ) cot ( −  ), Solution:,  5, , cos, −   = cos(450 0 −  ) = cos 360 + (90 0 −  ) = cos(90 0 −  ) = sin ,  2, , sin (4 +  ) = sin (2  2 +  ) = sin ,  5, , cos, − ,  2,  + tan (−  ) = sin  + − tan  = 1 + tan   tan , L.H .S =, sin (4 +  ) cot ( −  ) sin  − cot , , , , = 1 + tan 2  = sec 2  = R.H .S, , , , W .K .T, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , 1, = tan  and 1 + tan 2  = sec 2 , cot , , Page 21

Page 22 :
3.3.8 Problems on finding value of other ratio when when one of the ratio is given, Y, , 2, 1., If sin  = and  is acute, find tan  and sec , 3, 2 opp x, Solution: Given, sin  = =, =, 3 hyp r, From figure,, OA 2 = OB 2 + AB 2, , OB = OA − AB, 2, , 2, , 2, , x 2 = 32 − 2 2 = 9 − 4 = 5, , X/, , A, r=3, y=2, , , O, , x= 5, x= 5, , X, , B, , x, , Y/, , Since  lies in first quadrant , the value of x is positive, , Now,, Opp y, 2, = =, Adj x, 5, Hyp r, 3, sec  =, = =, Adj x, 5, , 1, 2., If sin  = and     is acute, find cos  , tan  and cos ec, 2, 2, A, 1 opp x, Solution: Given, sin  = =, =, 2 hyp r, From figure,, Opp=y=1, OA 2 = OB 2 + AB 2, tan  =, , OB 2 = OA 2 − AB 2, x = 2 −1 = 4 −1 = 3, 2, , 2, , 2, , X/, , Y, , Hyp=r=2, , , B, , Adj=x=?, , X, O, , x= 3, x=− 3, Now,, , Since  lies in II quadrant , the value of x is negative, , Y/, , Adj x − 3, 3, = =, =−, Hyp r, 2, 2, Opp y, 1, 1, tan  =, = =, =−, Adj x − 3, 3, Hyp r 2, cos ec =, = = =2, Opp y 1, 3 sin A + 2 cos A, 5, 3., If tan A =, and 180 0  A  270 0 then find the value of, 2 sin A − 3 cos A, 12, 5 opp y, Solution: Given, tan A =, =, =, B x=adj=-12, 12 Adj x, X/, Since A lies in III quadrant both x and y are negative, A,  x=-12 and y=-5, y=opp=-5, From figure,, cos  =, , Y, O, X, , R=hyp=?, A, Y/, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 22

Page 23 :
OA 2 = OB 2 + AB 2, r 2 = (− 12) + (− 5) = 144 + 25 = 169, 2, , 2, , r =  169 = 13, r = 13, , Since Hypotenuse is always positive, , Now,, , Opp y − 5, 5, = =, =−, Hyp r 13, 13, Adj x − 12, 12, cos A =, = =, =−, Hyp r, 13, 13, Consider given expression,  5   12 , 3 −  + 2 −  − 15 − 24 − 15 − 24, 3 sin A + 2 cos A, − 39, 3, 13   13 , 13, = , = 13 13 =, =, =−, 10, 36, −, 10, +, 36, 2 sin A − 3 cos A, 26, 2,  5   12 , 2 −  − 3 −  − +, 13 13, 13,  13   13 , 3 sin x − 2 cos x, 3, 13,  A  2 then find the value of, 4., If sec x =, and, 9 cos x + 4 sin x, 2, 5, 13 Hyp r, Y, Solution: Given, sec x =, =, =, 5, Adj x, O, x=adj=5, B, sin A =, , X/, , From figure,, OA 2 = OB 2 + AB 2, , x, , AB 2 = OA 2 − OB 2, , r=hyp=13, , y=opp=?, , y 2 = (13) − (5) = 144 − 25 = 144, 2, , 2, , y =  144 = 12, , X, , A, Y/, , y = −12 Since angle x lies in IV quadrant , y is negative, Now,, Opp y − 12, 12, sin x =, = =, =−, Hyp r, 13, 13, Adj x 5, cos x =, = =, Hyp r 13, ,  12   5 , 3 −  − 2  − 36 − 10 − 36 − 10, 3 sin x − 2 cos x, − 46 46, 13   13 , 13, = , = 13 13 =, =, =, 45 48, 45 − 48, 9 sin x + 4 cos x, −3, 3,  5   12 , −, 9  + 4 − , 13 13, 13,  13   13 , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 23

Page 24 :
3.4 COMPOUND ANGLES, 3.4, , 1., , 2., , Compound Angles: The sum or difference of angles like (A+B),(A-B),(A+B+C) and (A+BC) etc are called as compound angles, In this part of the topic we learn how to express trigonometric ratio of the sum(A+B) and, difference(A-B) in terms of ratios of angles A and B.[ without proof], Basically compound angles are classified in to two types they are Addition and Subtraction, fprmulae, Addition Formulae:, 1., sin ( A + B ) = sin A cos B + cos A sin B, 2., cos( A + B ) = cos A cos B − sin A sin B, tan A + tan B, 3., tan ( A + B ) =, 1 − tan A tan B, cot A cot B − 1, 4., cot ( A + B ) =, cot A + cot B, Subtraction Formulae:, 1., 2., , sin ( A − B ) = sin A cos B − cos A sin B, , cos( A − B ) = cos A cos B + sin A sin B, , tan A − tan B, 1 + tan A tan B, cot A cot B + 1, 4., cot ( A − B ) =, cot B − cot A, 3.4.1 Results on trigonometric ratios of an angles 150,750 and 1050 using compound angles, 3., , tan ( A − B ) =, , 1., Find the values of sin 150 , cos 150 and tan 150, Solution:, sin 150 = sin 450 − 30 0, , (, , ), , W .K .T sin ( A − B ) = sin A cos B − cos A sin B, sin 150 = sin 450 cos 30 0 − cos 450 sin 30 0, =, , 1, 3, 1 1, 3, 1, , −,  =, −, 2 2, 2 2 2 2 2 2, , 3 −1, 2 2, 0, cos 15 = cos 450 − 30 0, , sin 150 =, , (, , ), , W .K .T cos( A − B ) = cos A cos B + sin A sin B, cos 150 = cos 450 cos 30 0 + sin 450 sin 30 0, =, , 1, 3, 1 1, 3, 1, , +,  =, +, 2 2, 2 2 2 2 2 2, , 3 +1, 2 2, 0, tan 15 = tan 450 − 30 0, , cos 150 =, , (, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 24

Page 25 :
tan A − tan B, 1 + tan A tan B, 0, tan 45 − tan 30 0, tan 150 =, 1 + tan 450 tan 30 0, 1, 1, 3 −1, 1−, 1−, 3 =, 3 =, 3, =, 1, 1, 3 +1, 1 + 1, 1+, 3, 3, 3, , W .K .T tan ( A − B ) =, , =, , 3 −1, 3 −1, , =, , 3 −1, 3 −1, , 3 +1, 3 −1, , =, , (, , (, , ), , 2, , 3 −1, =, 3 +1 3 −1, , )(, , ), , ( 3), , + (1) − 2  3  1, 2, , ( 3 ) − (1), 3 2(2 − 3 ), =, 2, , 2, , 3 +1− 2 3 4 − 2, =, 3 −1, 2, 2, tan 150 = 2 − 3, Find the values of sin 750 , cos 750 and tan 750, =, , 2., , 2, , (, , sin 750 = sin 450 + 30 0, , ), , W .K .T sin ( A + B ) = sin A cos B + cos A sin B, sin 750 = sin 450 cos 30 0 + cos 450 sin 30 0, =, =, , 1, 3, 1 1, , +, , 2 2, 2 2, 3, 1, +, 2 2 2 2, , 3 +1, 2 2, 0, cos 75 = cos 450 + 30 0, , sin 750 =, , (, , ), , W .K .T cos( A + B ) = cos A cos B − sin A sin B, cos 750 = cos 450 cos 30 0 − sin 450 sin 30 0, =, =, , 1, 3, 1 1, , −, , 2 2, 2 2, 3, 1, −, 2 2 2 2, , 3 −1, 2 2, 0, tan 75 = tan 450 + 30 0, tan A + tan B, W .K .T tan ( A + B ) =, 1 − tan A tan B, 0, tan 45 + tan 30 0, tan 750 =, 1 − tan 450 tan 30 0, cos 750 =, , (, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 25

Page 26 :
1, 1+, 3, 0, tan 75 =, =, 1, 1 − 1, 1−, 3, 1+, , =, , 3 +1, 3 −1, , =, , 3 +1, 3 +1, , 3 −1, 3 +1, , =, , (, , (, , ), , 2, , 3 +1, =, 3 −1 3 +1, , )(, , ), , ( 3), , 3 +1+ 2 3 4 + 2, =, 3 −1, 2, 0, tan 75 = 2 + 3, =, , 3., , 3 +1, 3, 3 −1, 3, , 1, 3 =, 1, 3, , 2, , + (1) + 2  3  1, 2, , ( 3 ) − (1), 3 2(2 + 3 ), =, 2, , 2, , 2, , Find the values of sin 1050 , cos 1050 and tan 1050, , Solution:, , (, , sin 1050 = sin 60 0 + 450, , ), , W .K .T sin ( A + B ) = sin A cos B + cos A sin B, sin 1050 = sin 60 0 cos 450 + cos 60 0 sin 450, 3 1, 1 1, , + , 2, 2 2, 2, , =, =, , 3, 1, +, 2 2 2 2, 3 +1, 2 2, , sin 1050 =, , (, , cos 1050 = cos 60 0 + 450, , ), , W .K .T cos( A + B ) = cos A cos B − sin A sin B, cos 1050 = cos 60 0 cos 450 − sin 60 0 sin 450, =, =, , 1 1, 3 1, , −, , 2, 2, 2, 2, 1, 2 2, , cos 1050 =, , −, , 3, 2 2, , 1− 3, 2 2, , (, , tan 1050 = tan 60 0 + 450, , ), , tan A + tan B, 1 − tan A tan B, 0, tan 60 + tan 450, tan 1050 =, 1 − tan 60 0 tan 450, , W .K .T tan ( A + B ) =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 26

Page 27 :
tan 1050 =, , 3 +1, 3 +1, =−, 1 − 3 1, 3 −1, 3 +1, 3 +1, −, 3 −1, 3 +1, , =−, , (, , ), , ( 3 ) + (1) + 2  3  1, =−, ( )( ), ( 3 ) − (1), 3 +1+ 2 3, 4+2 3, 2(2 + 3 ), =−, =−, =−, 2, , 2, , 3 +1, =−, 3 −1 3 +1, , 3 −1, tan 105 = − 2 + 3, 0, , (, , ), , 2, , 2, , 2, , 2, , 2, , 3.4.2 Problems on finding the value of trigonometric ratios of compound angles, , 1., , If tan A =, , Solution:, , 1, 1, and tan B = , find the value of tan ( A + B) ., 2, 3, , Given tan A =, , 1, 1, and tan B =, 2, 3, , Consider, tan A + tan B, 1 − tan A tan B, 1 1, +, 2, 3, =, 1 1, 1− , 2 3, 1 1 3+ 2 5, +, 2, 3= 6 = 6, =, 1, 6 −1 5, 1−, 6, 6, 6, tan ( A + B ) = 1, , tan ( A + B ) =, , 2., , If tan A =, , Solution:, , 1, , 1, and tan B = , then prove that A + B = ., 3, 4, 2, , Given tan A =, , 1, 1, and tan B =, 3, 2, , Consider, tan A + tan B, 1 − tan A tan B, 1 1, +, = 3 2, 1 1, 1− , 3 2, , tan ( A + B ) =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 27

Page 28 :
1 1 2+3 5, +, 3, 2 = 6 = 6, =, 1, 6 −1 5, 1−, 6, 6, 6, tan ( A + B ) = 1,  tan ( A + B ) = tan,  A+ B =, , 3., , If tan A =, , Solution:, , , , , 4, , 4, , 4, , 1, and tan B = , then prove that A + B = ., 5, 4, 9, , Given tan A =, , 4, 1, and tan B =, 5, 9, , Consider, tan A + tan B, 1 − tan A tan B, 4 1, +, 5, 9, =, 4 1, 1− , 5 9, 36 + 5 41, = 45 = 45, 45 − 4 41, 45, 45, tan ( A + B ) = 1, , tan ( A + B ) =, ,  tan ( A + B ) = tan,  A+ B =, , 4., , If tan A =, , Given tan A =, , , , , 4, , 4, , 3, 1, and tan B = , then find A + B, 4, 7, , 3, 1, and tan B =, 4, 7, , Consider, tan A + tan B, 1 − tan A tan B, 3 1, 21 + 4 25, +, = 4 7 = 28 = 28, 3 1 28 − 3 25, 1− , 4 7, 28, 28, , tan ( A + B ) =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 28

Page 29 :
tan ( A + B ) = 1,  tan ( A + B ) = tan,  A+ B =, , 5., , If sin A =, , Solution:, , , , , 4, , 4, , 1, 1, , then show that A + B = 450, and sin B =, 10, 5, , Given sin A =, , 1, 1, and sin B =, 10, 5, , 1, 10, 2, cos A = 1 − sin 2 A, , 1, 5, 2, cos B = 1 − sin 2 B, , sin A =, ,  1 , =1− , ,  10 , 1, =1−, 10, 10 − 1, =, 10, 9, cos 2 A =, 10, 9, cos A =, 10, 3, cos A =, 10, , sin B =, , 2, ,  1 , =1− , ,  5, 1, =1−, 5, 5 −1, =, 5, 4, cos 2 B =, 5, 4, cos B =, 5, 2, cos B =, 5, , 2, , Consider, , sin ( A + B ) = sin A cos B + cos A sin B, 1, 2, 3, 1, , +, , 10, 5, 10, 5, 2, 3, =, +, 50, 50, 2+3, 5, 5, sin ( A + B ) =, =, =, 50, 25  2 5 2, 1, sin ( A + B ) =, 2,  sin ( A + B ) = sin 450, =, ,  A + B = 450, 6., , If sin A =, , 4, 5, and sin B = , then find cos( A + B) and cos( A − B), 5, 13, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 29

Page 30 :
Solution:, , Given sin A =, , 4, 5, and sin B = ,, 5, 13, , 4, 5, 2, cos A = 1 − sin 2 A, , 5, 13, 2, cos B = 1 − sin 2 B, , sin A =, , 4, =1−  , 5, 16, =1−, 25, 25 − 16, =, 25, 9, cos 2 A =, 25, 9, cos A =, 25, 3, cos A =, 5, , sin B =, , 2, , 2, , 5, =1−  ,  13 , 25, =1−, 169, 169 − 25, =, 169, 144, cos 2 B =, 169, 144, cos B =, 169, 12, cos B =, 13, , Consider, cos( A + B ) = cos A cos B − sin A sin B, 3 12 4 5,  − , 5 13 5 13, 36 20, =, −, =, 65 65, 36 − 20 16, cos( A + B ) =, =, 65, 65, =, , cos( A − B ) = cos A cos B + sin A sin B, 3 12 4 5,  + , 5 13 5 13, 36 20, =, +, =, 65 65, 36 + 20 56, cos( A − B ) =, =, 65, 65, =, , 3.4.3 Problems on identities using compound angle formulae, , 1., , Prove that sin ( A + B ) sin ( A − B ) = sin 2 A − sin 2 B, , Solution:, L.H .S = sin ( A + B )sin ( A − B ), = (sin A cos B + cos A sin B )(sin A cos B − cos A sin B ), = (sin A cos B ) − (cos A sin B ), 2, , 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 30

Page 31 :
= sin 2 A cos 2 B − cos 2 A sin 2 B, , (, , ) (, , ), , = sin 2 A 1 − sin 2 B − 1 − sin 2 A sin 2 B, = sin A − sin A sin B − sin B + sin 2 A sin 2 B, 2, , 2, , 2, , 2, , L.H .S = sin 2 A − sin 2 B = R.H .S, , 2., , Prove that sin ( A + B ) sin ( A − B ) = cos 2 B − cos 2 A, , Solution:, L.H .S = sin ( A + B ) sin ( A − B ), = (sin A cos B + cos A sin B )(sin A cos B − cos A sin B ), = (sin A cos B ) − (cos A sin B ), 2, , 2, , = sin 2 A cos 2 B − cos 2 A sin 2 B, , (, , ), , (, , = 1 − cos 2 A cos 2 B − cos 2 A 1 − cos 2 B, , ), , = cos 2 B − cos 2 A cos 2 B − cos 2 A + cos 2 A cos 2 B, L.H .S = cos 2 B − cos 2 A = R.h.S, , 3., , Prove that cos( A + B ) cos( A − B ) = cos 2 B − sin 2 A, , Solution:, L.H .S = cos( A + B ) cos( A − B ), , = (cos A cos B − sin A sin B )(cos A cos B + sin A sin B ), = (cos A cos B ) − (sin A sin B ), 2, , 2, , = cos 2 A cos 2 B − sin 2 A sin 2 B, , (, , ), , (, , = 1 − sin 2 A cos 2 B − sin 2 A 1 − cos 2 B, , ), , = cos B − sin A cos B − sin A + sin A cos 2 B, 2, , 2, , 2, , 2, , 2, , = cos 2 B − sin 2 A, = R.H .S, , 4., , Prove that cos( A + B ) cos( A − B ) = cos 2 A − sin 2 B, , Solution:, L.H .S = cos( A + B ) cos( A − B ), , = (cos A cos B − sin A sin B )(cos A cos B + sin A sin B ), = (cos A cos B ) − (sin A sin B ), 2, , 2, , = cos 2 A cos 2 B − sin 2 A sin 2 B, , (, , ) (, , ), , = cos 2 A 1 − sin 2 B − 1 − cos 2 A sin 2 B, = cos A − cos A sin B − sin B + cos 2 A sin 2 B, 2, , 2, , 2, , 2, , = cos 2 A − sin 2 B, = R.H .S, , 5., , Prove that, , sin ( A − B ) sin (B − C ) sin (C − A), +, +, =0, cos A cos B cos B cos C cos C cos A, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 31

Page 32 :
Solution:, L.H .S =, , sin ( A − B ) sin (B − C ) sin (C − A), +, +, cos A cos B cos B cos C cos C cos A, , sin ( A − B ), cos A cos B, sin A cos B − cos A sin B, =, cos A cos B,  sin A cos B cos A sin B , = , −, ,  cos A cos B cos A cos B , , =, ,  sin A sin B , = , −, ,  cos A cos B , =  (tan A − tan B ), = tan A − tan B + tan B − tan C + tan c − tan A, = 0 = R.H .S, , 6., , Prove that, , sin ( A − B ) sin (B − C ) sin (C − A), +, +, =0, sin A sin B sin A sin B sin A sin B, , Solution:, L.H .S =, , sin ( A − B ) sin (B − C ) sin (C − A), +, +, sin A sin B sin A sin B sin A sin B, , sin ( A − B ), sin A sin B, sin A cos B − cos A sin B, =, sin A sin B,  sin A cos B cos A sin B , = , −, ,  sin A sin B sin A sin B , , =, ,  cos B cos A , = , −, ,  sin B sin A , =  (cot B − cot A), = cot B − cot A + cot C − cot B + cot A − cot C, = 0 = R.H .S, , 7., , Prove that, , sin ( A + B ) tan A + tan B, =, sin ( A − B ) tan A − tan B, , Solution:, sin ( A + B ) sin A cos B + cos A sin B, =, sin ( A − B ) sin A cos B − cos A sin B, Divide both NR and DR by cos A cos B, , L.H .S =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 32

Page 33 :
sin A cos B cos A sin B, sin A sin B, +, +, cos, A, cos, B, cos, A, cos, B, cos, A, cos B, =, =, sin A cos B cos A sin B, sin A sin B, −, −, cos A cos B cos A cos B cos A cos B, tan A + tan B, =, = R.H .S, tan A − tan B, , 8., , Prove that, , sin ( A − B ) tan A − tan B, =, sin ( A + B ) tan A + tan B, , Solution:, sin ( A − B ) sin A cos B − cos A sin B, =, sin ( A + B ) sin A cos B + cos A sin B, Divide both NR and DR by cos A cos B, , L.H .S =, , sin A cos B cos A sin B, sin A sin B, −, −, cos, A, cos, B, cos, A, cos, B, cos, A, cos B, =, =, sin A cos B cos A sin B, sin A sin B, +, +, cos A cos B cos A cos B cos A cos B, tan A − tan B, =, = R.H .S, tan A + tan B, , 9., , Prove that, , cos( A + B ) + cos( A − B ), = cot A, sin ( A + B ) + sin ( A − B ), , Solution:, cos( A + B ) + cos( A − B ), sin ( A + B ) + sin ( A − B ), cos A cos B − sin A sin B + cos A cos B + sin A sin B, =, sin A cos B + cos A sin B + sin A cos B − cos A sin B, 2 cos A cos B cos A, =, =, 2 sin A cos B sin A, = cot A = R.H .S, , L.H .S =, , 10., , Prove that, , cos( A − B ) − cos( A + B ), = tan B, sin ( A − B ) + sin ( A + B ), , Solution:, cos( A − B ) − cos( A + B ), sin ( A − B ) + sin ( A + B ), cos A cos B + sin A sin B − (cos A cos B − sin A sin B ), =, sin A cos B − cos A sin B + sin A cos B + cos A sin B, cos A cos B + sin A sin B − cos A cos B + sin A sin B, =, sin A cos B − cos A sin B + sin A cos B + cos A sin B, 2 sin A sin B sin B, =, =, = tan B = R.H .S, 2 sin A cos B cos B, , L.H .S =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 33

Page 34 :
11., , Prove that, , sin ( A + B ) + sin ( A − B ), = tan A, cos( A + B ) + cos( A − B ), , Solution:, L.H .S =, , sin ( A + B ) + sin ( A − B ), cos( A + B ) + cos( A − B ), , sin A cos B + cos A sin B + sin A cos B − cos A sin B, cos A cos B − sin A sin B + cos A cos B + sin A sin B, 2 sin A cos B sin A, =, =, = tan A = R.H .S, 2 cos A cos B cos A, , L.H .S =, , 12., , Prove that, , sin ( A + B ) − sin ( A − B ), = tan B, cos( A + B ) + cos( A − B ), , Solution:, sin ( A + B ) − sin ( A − B ), cos( A + B ) + cos( A − B ), sin A cos B + cos A sin B − (sin A cos B − cos A sin B ), =, cos A cos B − sin A sin B + cos A cos B + sin A sin B, sin A cos B + cos A sin B − sin A cos B + cos A sin B, =, cos A cos B − sin A sin B + cos A cos B + sin A sin B, 2 cos A sin B sin B, =, =, = tan B = R.H .S, 2 cos A cos B cos B, , L.H .S =, , 13., , Prove that cot ( A + B ) =, , cot A cot B − 1, cot B + cot A, , Solution:, L.H .S = cot ( A + B ), , cos( A + B ), sin ( A + B ), cos A cos B − sin A sin B, =, sin A cos B + cos A sin B, Divide both NR and DR by sin A sin B, =, , cos A cos B sin A sin B, −, = sin A sin B sin A sin B, sin A cos B cos A sin B, +, sin A sin B sin A sin B, cot A cot B − 1, =, = R.H .S, cot B + cot A, , 14., , Prove that cot ( A − B ) =, , cot A cot B + 1, cot B − cot A, , Solution:, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 34

Page 35 :
L.H .S = cot ( A − B ), , cos( A − B ), sin ( A − B ), cos A cos B + sin A sin B, =, sin A cos B − cos A sin B, Divide both NR and DR sin A sin B, =, , cos A cos B sin A sin B, +, = sin A sin B sin A sin B, sin A cos B cos A sin B, −, sin A sin B sin A sin B, cot A cot B + 1, =, = R.H .S, cot B − cot A, , , , , , Prove that sin  A +  − cos A +  = 2 sin A, 4, 4, , , Solution:, 15., , , , , , L.H .S = sin  A +  − cos A + , 4, 4, , , , , , −  cos A cos − sin A sin , 4, 4 , 4, 4, 1, 1, 1, 1, = sin A , + CosA , − cos A , + sin A , 2, 2, 2, 2, sin A cos A cos A sin A, =, +, −, +, 2, 2, 2, 2, sin A sin A sin A + sin A, =, +, =, 2, 2, 2, , = sin A cos, , =, , 2 sin A, =, 2, , , , + cos A sin, , , , 2  2  sin A, 2, , = 2 sin A = R.H .S, , , , , , Prove that sin  A −  − cos A −  = − 2 cos A, 4, 4, , , Solution:, 16., , , , , , L.H .S = sin  A −  − cos A − , 4, 4, , , , , , −  cos A cos + sin A sin , 4, 4 , 4, 4, 1, 1, 1, 1, = sin A , − CosA , − cos A , − sin A , 2, 2, 2, 2, cos A cos A − cos A − cos A, =−, −, =, 2, 2, 2, , = sin A cos, , =, , , , − cos A sin, , , , − 2 cos A, 2  2  cos A, =−, 2, 2, , = − 2 cos A = R.H .S, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 35

Page 36 :
, , , , Prove that cos + A  + cos − A  = 2 cos A, 4, , 4, , Solution:, , , , , L.H .S = cos + A  + cos − A , 4, , 4, , , 17., , = cos, , , 4, , = 2 cos, , cos A − sin, , , , , 4, , sin A + cos, , , 4, , cos A + sin, , , 4, , sin A, , cos A, 4, 1, = 2,  cos A, 2, 1, = 2 2,  cos A, 2, = 2 cos A = R.H .S, ,  , , 18., Prove that tan  + A  tan  − A  = 1, 4,  4, , Solution:, ,  , , L.H .S = tan  + A  tan  − A , 4,  4, , =, , tan, , , 4, , 1 − tan, , + tan A, , , , , , tan, , , 4, , tan A 1 = tan, , − tan A, , , , tan A, 4, 4,  1 + tan A  1 − tan A , =, , ,  1 − 1  tan A  1 + 1  tan A ,  1 + tan A  1 − tan A , =, , ,  1 − tan A  1 + tan A , = 1 = R.H .S, sin 2 A cos 2 A, −, = sec A, 19., Prove that, sin A, cos A, Solution:, sin 2 A cos 2 A, L.H .S =, −, sin A, cos A, sin 2 A cos A − cos 2 A sin A, =, sin A cos A, , =, , sin (2 A − A), sin A, =, sin A cos A sin A cos A, , 1, = sec A = R.H .S, cos A, cos 2 A sin 2 A, +, = cos ecA, 20., Prove that, sin A, cos A, Solution:, =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 36

Page 37 :
cos 2 A sin 2 A, +, sin A, cos A, cos 2 A cos A + sin 2 A sin A, L.H .S =, sin A cos A, cos(2 A − A), =, sin A cos A, cos A, 1, =, =, sin A cos A sin A, = cos ecA = R.H .S, Prove that tan 5 − tan 3 − tan 2 = tan 5 tan 3 tan 2 ., Solution:, Consider, tan 5 = tan (3 + 2 ), L.H .S =, , 21., , tan 3 + tan 2, 1 − tan 3 tan 2, tan 5 (1 − tan 3 tan 2 ) = tan 3 + tan 2, , tan 5 =, , tan 5 − tan 5 tan 3 tan 2 = tan 3 + tan 2, tan 5 − tan 3 − tan 2 = tan 5 tan 3 tan 2, , 3.4.4 Problems on values of an expression using compound angle formulae, 1., Find the value of sin 40 0 cos 20 0 + cos 40 0 sin 20 0, Solution: Using sin A cos B + cos A sin B = sin ( A + B), Consider given expression, sin 40 0 cos 20 0 + cos 40 0 sin 20 0, , (, , = sin 40 0 + 20 0, = sin 60, =, , ), , 0, , 3, 2, , 2., Find the value of sin 100 0 cos 10 0 − cos 100 0 sin 10 0, Solution: Using sin A cos B − cos A sin B = sin ( A − B), Consider given expression, sin 100 0 cos 10 0 − cos 100 0 sin 10 0, , (, , = sin 100 0 − 10 0, = sin 90, , ), , 0, , =1, 3., Find the value of cos 750 cos 150 − sin 750 sin 150, Solution: Using cos A cos B − sin A sin B = cos( A + B), Consider given expression, , cos 750 cos 150 − sin 750 sin 150, , (, , ), , = cos 750 + 150 = cos 900 = 0, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 37

Page 38 :
4., Find the value of cos 140 0 cos 20 0 + sin 140 0 sin 20 0, Solution: Using cos A cos B + sin A sin B = cos( A − B), Consider given expression, cos 140 0 cos 20 0 + sin 140 0 sin 20 0, , (, , = cos 140 0 − 20 0, = cos 120 0, , (, , = cos 180 − 60 0, = − cos 60, =−, , 5., , Show that, , ), , ), , 0, , 1, 2, , tan 70 0 + tan 650, = −1, 1 − tan 70 0 tan 650, , Solution:, tan 70 0 + tan 650, 1 − tan 70 0 tan 650, tan A + tan B, w.k .t tan ( A + B ) =, 1 − tan A tan B, 0, = tan 70 + 650, L.H .S =, , (, , = tan 1350, , ), = tan (180 − 45 ), 0, , = − tan 450 = −1 = R.H .S, , 6., , Show that, , tan 1700 − tan 200, 1, =−, 0, 0, 1 + tan 170 tan 20, 3, , Solution:, tan 170 0 − tan 20 0, 1 + tan 170 0 tan 20 0, tan A − tan B, w.k .t tan ( A − B ) =, 1 + tan A tan B, 0, = tan 170 − 20 0, L.H .S =, , (, , = tan 150 0, , ), = tan (180 − 30 ), , = − tan 30 0 = −, , 0, , 1, = R.H .S, 3, , cos 150 + sin 150, 7., Show that, = 3, cos 150 − sin 150, Solution:, cos 150 + sin 150, L.H .S =, cos 150 − sin 150, Divede both NR DR by cos 150, cos 150 sin 150, +, 0, 0, cos, 15, cos 150 = 1 + tan 15, =, cos 150 sin 150 1 − tan 150, −, cos 150 cos 150, tan 450 + tan 150, =, = tan 450 + 150 = tan 60 0 = 3 = R.H .S, 0, 0, 1 − tan 45  tan 15, , (, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 38

Page 39 :
8., , Show that, , cos 17 0 + sin 17 0, = tan 620, cos 17 0 − sin 17 0, , Solution:, , cos 17 0 + sin 17 0, L.H .S =, cos 17 0 − sin 17 0, Divede both NR DR by cos 17 0, cos 17 0 sin 17 0, +, 0, 0, 1 + tan 17 0, = cos 17 0 cos 170 =, cos 17, sin 17, 1 − tan 17 0, −, cos 17 0 cos 17 0, tan 450 + tan 17 0, =, 1 − tan 450  tan 17 0, = tan 450 + 17 0, , (, , ), , = tan 62 = R.H .S, 3.4.5 Problems on conditional Identities using compound angle formulae, 0, , 1., , If A + B =, , , 4, , then prove that (1 + tan A)(1 + tan B) = 2 and hence deduce that, , 0, , 1, tan 22, = 2 −1, 2, Solution:, Given, , A+ B =, , , , 4, Taking tan on both sides, tan ( A + B ) = tan, , , 4, , tan A + tan B, =1, 1 − tan A tan B, tan A + tan B = 1(1 − tan A tan B ), tan A + tan B = 1 − tan A tan B, tan A + tan B + tan A tan B = 1, Add 1 on both sides, 1 + tan A + tan B + tan A tan B = 1 + 1, , (1 + tan A) + tan B(1 + tan A) = 2, (1 + tan A)(1 + tan B ) = 2, 0, ,  tan B is common, ,  (1 + tan A) is common, , 0, , 1, 1, and B = 22 so that A + B = 450 in the above equation, we get, 2, 2, 0, 0, , 1 , 1 , 1 + tan 22 1 + tan 22  = 2, , , , 2 , 2 , , , Taking A = 22, , 2, , 0, , 1 , 1 + tan 22  = 2, , 2 , , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 39

Page 40 :
0, , 1 + tan 22, , 1, = 2, 2, , 0, , 1, tan 22, = 2 −1, 2, , 2., , If A + B + C =, , , 2, , then prove that tan A tan B + tan B tan C + tan C tan A = 1, , Solution:, , Given, A+ B+C =, A+ B =, , , , , 2, , −C, 2, Taking tan on both side, , , tan ( A + B ) = tan  − C , 2, , tan A + tan B, = − cot C u sin g allied angle, 1 − tan A tan B, tan A + tan B, 1, =−, 1 − tan A tan B, tan C, By cross multiplica tion, , (tan A + tan B ) tan C = 1(1 − tan A tan B ), tan A tan C + tan B tan C = 1 − tan A tan B, tan A tan C + tan B tan C + tan A tan B = 1, tan A tan B + tan B tan C + tan C tan A = 1, 3., If A + B + C =  prove that tan 2 A + tan 2B + tan 2C = tan 2 A tan 2B tan 2C, Solution:, w.k .t sum of three angles of a ABC is 180 0 = , , A+ B+C =, A+ B = −C, Multiply both sides by 2, 2 A + 2 B = 2 − 2C, taking tan on both sides, , tan (2 A + 2 B ) = tan (2 − 2C ), , tan 2 A + tan 2 B, = − tan 2C, 1 − tan 2 A tan 2 B, tan 2 A + tan 2 B = − tan 2C (1 − tan 2 A tan 2 B ), tan 2 A + tan 2 B = − tan 2C + tan 2 A tan 2 B tan 2C, tan 2 A + tan 2 B + tan 2C = tan 2 A tan 2 B tan 2C, 4., In any triangle ABC prove that tan A + tan B + tan C = tan A tan B tan C, Solution:, w.k .t sum of three angles of a ABC is 1800 = , A+ B +C =, A+ B = −C, taking tan on both sides, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 40

Page 41 :
tan ( A + B ) = tan ( − C ), tan A + tan B, = − tan C, 1 − tan A tan B, tan A + tan B = − tan C (1 − tan A tan B ), tan A + tan B = − tan C + tan A tan B tan C, tan A + tan B + tan C = tan A tan B tan C, , Multiples and submultiples Angles, 3.5, , Multiples and Submultiples Angles:, If A is an angle then the angles of the form 2A, 3A, 4A etc are called multiple angles of A, A A, and , etc are called submultiples angles of A, 2 3, 3.5.1 Trigonometric ratios of multiple angles:, 1., Prove that sin2A=2sinAcosA, Proof:, Consider, sin ( A + B ) = sin A cos B + cos A sin B, put B = A, sin ( A + A) = sin A cos A + cos A sin A, sin 2 A = 2 sin A cos A, Hence proved the result ., 2., Proof:, , Prove that sin 2 A =, , 2 tan A, ., 1 + tan 2 A, , Consider, W .K .T, sin 2 A = 2 sin A cos A, 2 sin A cos A, sin 2 A =, 1, 2 sin A cos A, sin 2 A =, cos 2 A + sin 2 A, 2 sin A cos A, sin 2 A =, , sin 2 A , 2, , cos A1 +, 2,  cos A , , By taking 1 = cos 2 A + sin 2 A, By taking cos 2 A is common in DR, , sin A, cos A, sin 2 A =, 1 + tan 2 A, 2 tan A, sin 2 A =, 1 + tan 2 A, Hence proved the result ., 2, , 3., Prove that cos 2 A = cos 2 A − sin 2 A., Proof:, consider, cos( A + B ) = cos A cos B − sin A sin B, put B = A, cos( A + A) = cos A cos A − sin A sin A, cos 2 A = cos 2 A − sin 2 A., Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 41

Page 42 :
4., Prove that cos 2 A = 2 cos 2 A − 1., Proof:, consider, cos( A + B ) = cos A cos B − sin A sin B, put B = A, cos( A + A) = cos A cos A − sin A sin A, cos 2 A = cos 2 A − sin 2 A., , (, , cos 2 A = cos 2 A − 1 − cos 2 A, , ), , cos 2 A = cos A − 1 + cos A, 2, , 2, , cos 2 A = 2 cos 2 A − 1, Hence proved the result ., , 5., Prove that cos 2 A = 1 − 2 sin 2 A., Proof:, consider, cos( A + B) = cos A cos B − sin Asin B, put B = A, cos( A + A) = cos A cos A − sin A sin A, cos 2 A = cos 2 A − sin 2 A., cos 2 A = 1 − sin 2 A − sin 2 A, cos 2 A = 1 − 2 sin 2 A, Hence proved the result ., , 6., Proof:, , Prove that cos 2 A =, , 1 − tan 2 A, ., 1 + tan 2 A, , consider, cos 2 A = cos 2 A − sin 2 A, cos 2 A − sin 2 A, 1, 2, cos A − sin 2 A, cos 2 A =, cos 2 A + sin 2 A, , sin 2 A , , cos 2 A1 −, cos 2 A , , cos 2 A =, , sin 2 A , , cos 2 A1 +, 2,  cos A , 1 − tan 2 A, cos 2 A =, 1 + tan 2 A, Hence proved the result ., , cos 2 A =, , 7., , Proof:, , Prove that tan 2 A =, , 2 tan A, ., 1 − tan 2 A, , Consider compound angle formula, tan A + tan B, tan ( A + B ) =, 1 − tan A tan B, Put B = A, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 42

Page 43 :
tan A + tan A, 1 − tan A tan A, 2 tan A, tan 2 A =, 1 − tan 2 A, Hence proved the result, , tan ( A + A) =, , 8., Prove that sin 3 A = 3 sin A − 4 sin 3 A., Proof:, Consider, sin 3 A = sin (2 A + A), , sin ( A + B ) = sin A cos B + cos A sin B, , sin 3 A = sin 2 A cos A + cos 2 A sin A, , (, , sin 2 A = 2 sin A cos A, , ), , sin 3 A = 2 sin A cos A  cos A + 1 − 2 sin A sin A cos 2 A = 1 − 2 sin 2 A, sin 3 A = 2 sin A cos 2 A + sin A − 2 sin 3 A, cos 2 A = 1 − sin 2 A, , (, , 2, , ), , sin 3 A = 2 sin A 1 − sin 2 A + sin A − 2 sin 3 A, sin 3 A = 2 sin A − 2 sin 3 A + sin A − 2 sin 3 A, sin 3 A = 3 sin A − 4 sin 3 A, Hence proved the result, , 9., Prove that cos 3 A = 4 cos 3 A − 3 cos A., Proof:, Consider, cos 3 A = cos(2 A + A), cos 3 A = cos 2 A cos A − sin 2 A sin A, , (, , ), , cos 3 A = 2 cos 2 A − 1 cos A − 2 sin A cos A  sin A, cos 3 A = 2 cos A − cos A − 2 sin 2 A  cos A, 3, , (, , cos 3 A = 2 cos 3 A − cos A − 2 cos A 1 − cos 2 A, , ), , cos 3 A = 2 cos 3 A − cos A − 2 cos A + 2 cos 3 A, cos 3 A = 4 cos 3 A − 3 cos A, Hence proved the result, , 10., Proof:, , Prove that tan 3 A =, , 3 tan A − tan 3 A, 1 − 3 tan 2 A, , Consider, tan 3 A = tan (2 A + A), , tan 2 A + tan A, 1 − tan 2 A tan A, 2 tan A, + tan A, 2, tan 3 A = 1 − tan A,  2 tan A , 1− ,  tan A, 2,  1 − tan A , 2 tan A + tan A 1 − tan 2 A, 1 − tan 2 A, tan 3 A =, 1 − tan 2 A − 2 tan 2 A, 1 − tan 2 A, 2 tan A + tan A − tan 3 A, tan 3 A =, 1 − 3 tan 2 A, , tan 3 A =, , (, , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 43

Page 44 :
3 tan A − tan 3 A, 1 − 3 tan 2 A, Hence proved the result, 3.5.2 Trigonometric ratios of Submultiples angles:, tan 3 A =, , In the above results, if we take 2 A =   A =, , , 2, , then we get expressions for trigonometric, , ratios in terms of submultiples angle., , , 1. sin  = 2 sin cos, 2, 2, 2 tan, , 2. sin  =, , , 2, , , , 1 + tan 2, , 3. cos  = cos 2, , , , 2, − sin 2, , 2, , 4. cos  = 1 − 2 sin 2, 5. cos  = 2 cos 2, , 6. cos  =, , , , 1 + tan, 2 tan, , 7. tan  =, , 2, , 2, , , 2, −1, , 2, , 1 − tan 2, , , , , 2, , , , 2, , , 2, , 1 − tan 2, , , 2, , LIST OF FORMULAE, MULTIPLE ANGLE, 1. sin 2 A = 2 sin A cos A, , 2. sin 2 A =, , 2 tan A, ., 1 + tan 2 A, , SUBMULTIPLE ANGLE, 1. sin  = 2 sin, , 2. sin  =, , , 2, , 2 tan, , cos, , 4. cos 2 A = 2 cos 2 A − 1., 5. cos 2 A = 1 − 2 sin 2 A., , 3. cos  = cos 2, , , 2, , 4. cos  = 2 cos 2, , 2, , , 2, , − sin 2, , , 2, , 5. cos  = 1 − 2 sin 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , 2, , , , 1 + tan 2, , 3. cos 2 A = cos 2 A − sin 2 A., , , , , 2, , −1, , , 2, , Page 44

Page 45 :
6. cos 2 A =, , 7. tan 2 A =, , 1 − tan 2 A, ., 1 + tan 2 A, , 6. cos  =, , 1 − tan 2, 1 + tan, , 2 tan A, ., 1 − tan 2 A, , 7. tan  =, , 2 tan, 1 − tan, , 8. 1 + cos 2 A = 2 cos 2 A, , 2, , , 2, , , , 2, , , 2, 2, , , 2, , 8. 1 + cos  = 2 cos 2, , 9. 1 − cos 2 A = 2 sin 2 A., , 9. 1 − cos  = 2 sin 2, , , 2, , , 2, , 10. sin 3 A = 3 sin A − 4 sin 3 A., 11. cos 3 A = 4 cos 3 A − 3 cos A., 3 tan A − tan 3 A, 12. tan 3 A =, 1 − 3 tan 2 A, , 3.5.2 Problems on proof of identities using multiple angle formulae, 1., , Prove that (cos  + sin  ) = 1 + sin 2 ., 2, , L.H .S = (cos  + sin  ), , 2, , Solution:, , W .K .T (a + b ) = a 2 + b 2 + 2ab, 2, , L.H .S = cos 2  + sin 2  + 2 cos  sin , W .K .T cos 2  + sin 2  = 1 & 2 sin  cos  = sin 2, L.H .S = 1 + sin 2 = R.H .S, 2., , Prove that (cos  − sin  ) = 1 − sin 2 ., , Solution:, , 2, , L.H .S = (cos  − sin  ), , 2, , W .K .T (a − b ) = a 2 + b 2 − 2ab, 2, , L.H .S = cos 2  + sin 2  − 2 cos  sin , W .K .T cos 2  + sin 2  = 1 & 2 sin  cos  = sin 2, L.H .S = 1 − sin 2 = R.H .S, 1 − cos 2 A, = tan 2 A., 3., Prove that, 1 + cos 2 A, 1 − cos 2 A, L.H .S =, Solution:, 1 + cos 2 A, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 45

Page 46 :
W .K .T 1 − cos 2 A = 2 sin 2 A, 1 + cos 2 A = 2 cos 2 A, 2 sin 2 A, = tan 2 A = R.H .S, 2, 2 cos A, 1 + cos 2 A, = cot 2 A., 4., Prove that, 1 − cos 2 A, 1 + cos 2 A, L.H .S =, Solution:, 1 − cos 2 A, L.H .S =, , W .K .T 1 − cos 2 A = 2 sin 2 A, 1 + cos 2 A = 2 cos 2 A, L.H .S =, 5., , Prove that, , Solution:, , 2 cos 2 A, = cot 2 A = R.H .S, 2, 2 sin A, , 1 + cos 2 A, = cot A., 1 − cos 2 A, , L.H .S =, , 1 + cos 2 A, 1 − cos 2 A, , W .K .T 1 − cos 2 A = 2 sin 2 A, 1 + cos 2 A = 2 cos 2 A, L.H .S =, , 6., , Prove that, , Solution:, , 2 cos 2 A, = cot 2 A = cot A = R.H .S, 2, 2 sin A, , 1 − cos 2 A, = tan A., 1 + cos 2 A, , L.H .S =, , 1 − cos 2 A, 1 + cos 2 A, , W .K .T 1 − cos 2 A = 2 sin 2 A, 1 + cos 2 A = 2 cos 2 A, 2 sin 2 A, = tan 2 A = tan A = R.H .S, 2, 2 cos A, sin 2 A, = tan A., 7., Prove that, 1 + cos 2 A, sin 2 A, L.H .S =, Solution:, 1 + cos 2 A, W .K .T sin 2 A = 2 sin A cos A, L.H .S =, , 1 + cos 2 A = 2 cos 2 A, 2 sin A cos A, 2 cos 2 A, sin A, L.H .S =, = tan A = R.H .S, cos A, sin 2 A, = cot A., Prove that, 1 − cos 2 A, L.H .S =, , 8., , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 46

Page 47 :
Solution:, , sin 2 A, 1 − cos 2 A, W .K .T sin 2 A = 2 sin A cos A, L.H .S =, , 1 − cos 2 A = 2 sin 2 A, 2 sin A cos A, 2 sin 2 A, cos A, L.H .S =, = cot A = R.H .S, sin A, 1 + cos 2 A, = cot A., 9., Prove that, sin 2 A, 1 + cos 2 A, L.H .S =, Solution:, sin 2 A, W .K .T sin 2 A = 2 sin A cos A, L.H .S =, , 1 + cos 2 A = 2 cos 2 A, , 2 cos 2 A, 2 sin A cos A, cos A, L.H .S =, = cot A = R.H .S, sin A, 1 − cos 2 A, = tan A., 10., Prove that, sin 2 A, 1 − cos 2 A, L.H .S =, Solution:, sin 2 A, W .K .T sin 2 A = 2 sin A cos A, L.H .S =, , 1 − cos 2 A = 2 sin 2 A, , 2 sin 2 A, 2 sin A cos A, sin A, L.H .S =, = tan A = R.H .S, cos A, , L.H .S =, , 11., Prove that cos 4  − sin 4  = cos 2, L.H .S = cos 4  − sin 4 , Solution:, , (, , L.H .S = cos 2 , , ) − (sin  ), 2, , 2, , 2, , WKT a 2 − b 2 = (a + b )(a − b ), , (, , )(, , L.H .S = cos 2  + sin 2  cos 2  − sin 2 , , ), , WKT cos  + sin  = 1 & cos  − sin  = cos 2, 2, , 2, , 2, , 2, , L.H .S = (1)(cos 2 ) = cos 2 = R.H .S, , 12., , sin 3 A cos 3 A, −, =2, sin A, cos A, sin 3 A cos 3 A, L.H .S =, −, sin A cos A, , Prove that, , Solution:, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 47

Page 48 :
taking LCM, sin 3 A cos A − cos 3 A sin A, sin A cos A, Use sin A cos B − cos A sin B = sin ( A − B ) formula, L.H .S =, , sin (3 A − A), sin 2 A, 2 sin A cos A, =, =, sin A cos A sin A cos A, sin A cos A, L.H .S = 2 = R.H .S, , L.H .S =, , 13., , sin 2 A cos 2 A, −, = sec A, sin A, cos A, sin 2 A cos 2 A, L.H .S =, −, sin A, cos A, taking LCM, , Prove that, , Solution:, , sin 2 A cos A − cos 2 A sin A, sin A cos A, Use sin A cos B − cos A sin B = sin ( A − B ) formula, L.H .S =, , sin (2 A − A), sin A, 1, =, =, sin A cos A sin A cos A cos A, L.H .S = sec A = R.H .S, cos 3 A sin 3 A, −, = −2, 14., Prove that, cos A, sin A, cos 3 A sin 3 A, L.H .S =, −, Solution:, cos A sin A, taking LCM, L.H .S =, , cos 3 A sin A − sin 3 A cos A, sin A cos A, − (sin 3 A cos A − cos 3 A sin A), L.H .S =, sin A cos A, Use sin A cos B − cos A sin B = sin ( A − B ) formula, L.H .S =, , − sin (3 A − A), − sin 2 A, − 2 sin A cos A, =, =, sin A cos A, sin A cos A, sin A cos A, L.H .S = −2 = R.H .S, OR, cos 3 A sin 3 A, L.H .S =, −, cos A sin A, WKT cos 3 A = 4 cos 3 A − 3 cos A & sin 3 A = 3 sin A − 4 sin 3 A, L.H .S =, , L.H .S =, , 4 cos 3 A − 3 cos A 3 sin A − 4 sin 3 A, −, cos A, sin A, , (, , ), , (, , cos A 4 cos 2 A − 3 sin A 3 − 4 sin 2 A, −, cos A, sin A, 2, 2, L.H .S = 4 cos A − 3 − 3 − 4 sin A, , L.H .S =, , (, , ), , ), , L.H .S = 4 cos 2 A − 3 − 3 + 4 sin 2 A = 4 cos 2 A + 4 sin 2 A − 6, , (, , ), , L.H .S = 4 cos 2 A + sin 2 A − 6 = 4(1) − 6 = −2 = R.H .S, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 48

Page 49 :
15., , Prove that, , Solution:, , cos 3 A − cos 3 A sin 3 A + sin 3 A, +, = 3., cos A, sin A, , cos 3 A − cos 3 A sin 3 A + sin 3 A, +, cos A, sin A, 3, WKT cos 3 A = 4 cos A − 3 cos A & sin 3 A = 3 sin A − 4 sin 3 A, L.H .S =, , (, , ), , cos 3 A − 4 cos 3 A − 3 cos A sin 3 A + 3 sin A − 4 sin 3 A, +, cos A, sin A, 3, 3, 3, cos A − 4 cos A + 3 cos A sin A + 3 sin A − 4 sin 3 A, L.H .S =, +, cos A, sin A, , L.H .S =, , − 3 cos 3 A + 3 cos A 3 sin A − 3 sin 3 A, L.H .S =, +, cos A, sin A, 2, cos A − 3 cos A + 3 sin A 3 − 3 sin 2 A, L.H .S =, +, cos A, sin A, 2, 2, L.H .S = −3 cos A + 3 + 3 − 3 sin A, , (, , ), , (, , ), , L.H .S = 6 − 3 cos 2 A − 3 sin 2 A, , (, , L.H .S = 6 − 3 cos 2 A + sin 2 A, , ), , L.H .S = 6 − 3(1) = 6 − 3 = 3 = R.H .S, 16., , cos 3 A + cos 3 A sin 3 A − sin 3 A, Prove that, −, = −1, cos A, sin A, , Solution:, , L.H .S =, , cos 3 A + cos 3 A sin 3 A − sin 3 A, −, cos A, sin A, , WKT cos 3 A = 4 cos 3 A − 3 cos A & sin 3 A = 3 sin A − 4 sin 3 A, , (, , ), , cos 3 A + 4 cos 3 A − 3 cos A 3 sin A − 4 sin 3 A − sin 3 A, −, cos A, sin A, 3, 3, 5 cos A − 3 cos A 3 sin A − 5 sin A, L.H .S =, −, cos A, sin A, 2, cos A 5 cos A − 3 sin A 3 − 5 sin 2 A, L.H .S =, −, cos A, sin A, 2, 2, L.H .S = 5 cos A − 3 − 3 − 5 sin A, , L.H .S =, , (, , ), , (, , (, , ), , ), , L.H .S = 5 cos 2 A − 3 − 3 + 5 sin 2 A, , (, , ), , L.H .S = 5 cos 2 A + sin 2 A − 6, L.H .S = 5(1) − 6 = −1, , 17., , 1 − cos 2 A + sin 2 A, = tan A., 1 + cos 2 A + Sin 2 A, 1 − cos 2 A + sin 2 A, L.H .S =, 1 + cos 2 A + Sin 2 A, , Prove that, , Solution:, , WKT 1 − cos 2 A = 2 sin 2 A, 1 + cos 2 A = 2 cos 2 A, sin 2 A = 2 sin A cos A, L.H .S =, , 2 sin 2 A + 2 sin A cos A, 2 cos 2 A + 2 sin A cos A, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 49

Page 50 :
2 sin A(sin A + cos A), 2 cos A(cos A + sin A), sin A, L.H .S =, = tan A = R.H .S, cos A, 1 + cos 2 A + sin 2 A, = cot A., 18., Prove that, 1 − cos 2 A + Sin 2 A, 1 + cos 2 A + sin 2 A, L.H .S =, Solution:, 1 − cos 2 A + Sin 2 A, WKT 1 − cos 2 A = 2 sin 2 A, L.H .S =, , 1 + cos 2 A = 2 cos 2 A, sin 2 A = 2 sin A cos A, 2 cos 2 A + 2 sin A cos A, L.H .S =, 2 sin 2 A + 2 sin A cos A, 2 cos A(cos A + sin A), L.H .S =, 2 sin A(sin A + cos A), cos A, L.H .S =, = cot A = R.H .S, sin A, sin A + sin 2 A, = tan A., 19., Prove that, 1 + cos A + cos 2 A, sin A + sin 2 A, L.H .S =, Solution:, 1 + cos 2 A + cos A, WKT 1 + cos 2 A = 2 cos 2 A, sin 2 A = 2 sin A cos A, sin A + 2 sin A cos A, 2 cos 2 A + cos A, sin A(1 + 2 cos A), L.H .S =, cos A(2 cos A + 1), sin A, L.H .S =, = tan A = R.H .S, cos A, , L.H .S =, , 20., , cos A − sin A cos A + sin A, +, = 2 sec 2 A., cos A + SinA cos A − SinA, cos A − sin A cos A + sin A, L.H .S =, +, cos A + SinA cos A − SinA, , Prove that, , Solution:, , 2, 2, (, cos A − sin A) + (cos A + sin A), L.H .S =, (cos A + sin A)(cos A − sin A), , cos 2 A + sin 2 A − 2 cos A sin A + cos 2 A + sin 2 A + 2 cos A sin A, cos 2 A − sin 2 A, 1+1, 2, L.H .S =, =, = 2 sec 2 A, cos 2 A cos 2 A, cot 3 x + cot x, = −2 cos 2 x., Prove that, cot 3 x − cot x, L.H .S =, , 21., , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 50

Page 51 :
cot 3 x + cot x, cot 3 x − cot x, cos 3 x cos x cos 3 x sin x + sin 3 x cos x, sin 3 x cos x + cos 3 x sin x, +, sin 3 x sin x, sin 3 x sin x, L.H .S = sin 3 x sin x =, =, cos 3 x cos x cos 3x sin x − sin 3 x cos x − (sin 3 x cos x − cos 3 x sin x ), −, sin 3 x sin x, sin 3 x sin x, sin 3 x sin x, sin (3 x + x ), sin 4 x, sin 2(2 x ), 2 sin 2 x cos 2 x, L.H .S =, =−, =−, =−, = −2 cos 2 x, − sin (3 x − x ), sin 2 x, sin 2 x, sin 2 x, , L.H .S =, , Solution:, , 22., , Prove that tan  + cot  = 2 cos ec2 ., , Solution:, , L.H.S = tan  + cot , sin 2 = 2 sin  cos , , sin  cos  sin 2  + cos 2 , sin 2, L.H .S =, +, =,  sin  cos  =, 2, cos  sin , cos  sin , 1, 1, 2, L.H .S =, =, =, = 2 cos ec 2 = R.H .S, sin  cos  sin 2 sin 2, 2, , 23., , Prove that cos 4 A = 1 − 8 cos 2 A + 8 cos 4 A., , Solution:, , L.H .S = cos 4 A = cos 2(2 A), L.H .S = 2 cos 2 2 A − 1, L.H .S = 2(cos 2 A) − 1, 2, , L.H .S = 2(2 cos 2 A − 1) − 1, 2, , (, , ), , L.H .S = 2 (2 cos 2 A) + (1) − 2  2 cos 2 A  1 − 1, 2, , 2, , L.H .S = 2(4 cos A + 1 − 4 cos A) − 1, 4, , 2, , L.H .S = 8 cos 4 A + 2 − 8 cos 2 A − 1, L.H .S = 1 − 8 cos 2 A + 8 cos 4 A = R.H .S, 3.5.3 Problems on proof of identities using submultiples angle formulae, 2, , 1., , A, A, , Prove that  cos − sin  = 1 − sin A., 2, 2, , 2, , Solution:, , A, A, , L.H .S =  cos − sin , 2, 2, , A, A, A, A, L.H .S = cos 2 + sin 2 − 2 cos sin, 2, 2, 2, 2, A, A, L.H .S = 1 − 2 sin cos, 2, 2, L.H .S = 1 − sin A = R.H .S, 2, , A, A, , 2., Prove that  cos + sin  = 1 + sin A., 2, 2, , Solution:, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 51

Page 52 :
A, A, , L.H .S =  cos + sin , 2, 2, , , 2, , A, A, A, A, + sin 2 + 2 cos sin, 2, 2, 2, 2, A, A, L.H .S = 1 + 2 sin cos, 2, 2, L.H .S = 1 + sin A = R.H .S, , L.H .S = cos 2, , sin A, A, = cot ., 1 − cos A, 2, sin A, L.H .S =, Solution:, 1 − cos A, A, WKT sin A = 2 sin cos, 2, A, A, 2 sin cos, 2, 2 =, L.H .S =, A, 2 sin 2, 2, sin A, A, = tan ., 4., Prove that, 1 + cos A, 2, sin A, L.H .S =, Solution:, 1 + cos A, , 3., , Prove that, , A, A, & 1 − cos A = 2 sin 2, 2, 2, A, cos, 2 = cot A = R.H .S, A, 2, sin, 2, , A, A, cos, 2, 2, A, 1 + cos A = 2 cos 2, 2, A, A, A, 2 sin cos, sin, 2, 2 =, 2 = tan A = R.H .S, L.H .S =, A, A, 2, 2 cos 2, cos, 2, 2, , WKT sin A = 2 sin, , 5., , Solution:, , 6., , 1 + cos A, A, = cot ., sin A, 2, 1 + cos A, L.H .S =, sin A, A, A, WKT sin A = 2 sin cos, 2, 2, A, 1 + cos A = 2 cos 2, 2, A, A, 2 cos 2, cos, 2 =, 2 = cot A = R.H .S, L.H .S =, A, A, A, 2, 2 sin cos, sin, 2, 2, 2, , Prove that, , Prove that, , 1 − cos A, A, = tan ., sin A, 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 52

Page 53 :
Solution:, , L.H .S =, , 1 − cos A, sin A, , A, A, A, cos & 1 − cos A = 2 sin 2, 2, 2, 2, A, A, 2 sin 2, sin, 2 =, 2 = tan A = R.H .S, L.H .S =, A, A, A, 2, 2 sin cos, cos, 2, 2, 2, , WKT sin A = 2 sin, , 7., , 1 − cos x, x, = tan 2 ., 1 + cos x, 2, 1 − cos x, L.H .S =, 1 + cos x, , Prove that, , Solution:, , WKT 1 − cos x = 2 sin 2, , x, 2, , x, 2, x, x, 2 sin 2, sin 2, 2 =, 2 = tan 2 x = R.H .S, L.H .S =, x, x, 2, 2 cos 2, cos 2, 2, 2, 1 + cos x = 2 cos 2, , 8., , Prove that, , Solution:, , 1 − cos x, x, = tan ., 1 + cos x, 2, , L.H .S =, , 1 − cos x, 1 + cos x, , WKT 1 − cos x = 2 sin 2, , x, 2, , x, 2, x, x, 2 sin 2, sin, 2 =, 2 = tan x = R.H .S, L.H .S =, x, x, 2, 2 cos 2, cos, 2, 2, 1 + cos x, x, = cot 2 ., 9., Prove that, 1 − cos x, 2, 1 + cos x, L.H .S =, Solution:, 1 − cos x, x, WKT 1 − cos x = 2 sin 2, 2, x, 1 + cos x = 2 cos 2, 2, 1 + cos x = 2 cos 2, , 10., , Prove that, , Solution:, , 1 + cos x, x, = cot ., 1 − cos x, 2, , L.H .S =, , 1 + cos x, 1 − cos x, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 53

Page 54 :
WKT 1 − cos x = 2 sin 2, , x, x, & 1 + cos x = 2 cos 2, 2, 2, , x, x, cos, 2 =, 2 = cot x = R.H .S, L.H .S =, x, x, 2, 2 sin 2, sin, 2, 2, 1 − cos A + sin A, A, = tan ., Prove that, 1 + cos A + sin A, 2, 2 cos 2, , 11., , Solution:, , L.H .S =, , 1 − cos A + sin A, 1 + cos A + sin A, , WKT 1 − cos A = 2 sin 2, , A, 2, , A, 2, A, A, sin A = 2 sin cos, 2, 2, , 1 + cos A = 2 cos 2, , A, A, A, + 2 sin cos, 2, 2, 2, L.H .S =, A, A, A, 2 cos 2 + 2 sin cos, 2, 2, 2, A, A, A, 2 sin  sin + cos  sin A, 2, 2, 2, 2 = tan A = R.H .S, L.H .S =, =, A, A, A, A, 2, 2 cos  cos + sin  cos, 2, 2, 2, 2, 2 sin 2, , 12., , Prove that, , Solution:, , 1 + cos A + sin A, A, = cot ., 1 − cos A + sin A, 2, , L.H .S =, , 1 + cos A + sin A, 1 − cos A + sin A, , WKT 1 − cos A = 2 sin 2, , A, 2, , A, 2, A, A, sin A = 2 sin cos, 2, 2, , 1 + cos A = 2 cos 2, , A, A, A, + 2 sin cos, 2, 2, 2, L.H .S =, A, A, 2 A, 2 sin, + 2 sin cos, 2, 2, 2, A, A, A, 2 cos  cos + sin  cos A, 2, 2, 2, 2 = cot A = R.H .S, L.H .S =, =, A, A, A, A, 2, 2 sin  sin + cos  sin, 2, 2, 2, 2, 2 cos 2, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 54

Page 55 :
13., , Prove that cos ecA + cot A = cot, , Solution:, , A, ., 2, , L.H.S = cos ecA + cot A, 1, cos A, +, sin A sin A, 1 + cos A, LH .S =, sin A, , L.H .S =, , WKT 1 + cos A = 2 cos 2, , A, 2, , sin A = 2 sin, , A, A, cos, 2, 2, , A, 2, , cos, , A, 2 = cot A = R.H .S, L.H .S =, =, A, A, A, 2, 2 sin cos, sin, 2, 2, 2, 2 cos 2, , TRANSFORMATION FORMULAE, 3.6, , TRANSFORMATION FORMULAE:, , 3.6.1 Transformation formulae to express product of ratios into sum or difference of ratios:, Consider compound angle formulae, sin ( A + B ) = sin A cos B + cos A sin B, , − − − − −(1), , sin ( A − B ) = sin A cos B − cos A sin B, , − − − − −(2), , cos( A + B ) = cos A cos B − sin A sin B, , − − − − −(3), , cos( A − B ) = cos A cos B + sin A sin B, , − − − − −(4), , Equation (1) + Equation (2) , , sin ( A + B ) + sin ( A − B ) = sin A cos B + cos A sin B + sin A cos B − cos A sin B, sin ( A + B ) + sin ( A − B ) = 2 sin A cos B, , 2 sin A cos B = sin ( A + B ) + sin ( A − B ), , 2 sin A cos B = sin ( A + B) + sin ( A − B) Or, , sin A cos B =, , 1, sin ( A + B ) + sin ( A − B ), 2, , Equation (1) − Equation (2) , , sin ( A + B ) − sin ( A − B ) = sin A cos B + cos A sin B − (sin A cos B − cos A sin B ), sin ( A + B ) − sin ( A − B ) = sin A cos B + cos A sin B − sin A cos B + cos A sin B, sin ( A + B ) − sin ( A − B ) = 2 cos A sin B, , 2 cos A sin B = sin ( A + B ) − sin ( A − B ), , 2 cos A sin B = sin ( A + B) − sin ( A − B), , Or, , cos A sin B =, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , 1, sin ( A + B ) − sin ( A − B ), 2, , Page 55

Page 56 :
Equation (3) + Equation (4) , , cos( A + B ) + cos( A − B ) = cos A cos B − sin A sin B + cos A cos B + sin A sin B, cos( A + B ) + cos( A − B ) = 2 cos A cos B, , 2 cos A cos B = cos( A + B ) + cos( A − B ), , 2 cos A cos B = cos( A + B) + cos( A − B), , Or, , cos A cos B =, , 1, cos( A + B ) + cos( A − B ), 2, , Equation (3) − Equation (4) , , cos( A + B ) − cos( A − B ) = cos A cos B − sin A sin B − (cos A cos B + sin A sin B), cos( A + B ) − cos( A − B ) = cos A cos B − sin A sin B − cos A cos B − sin A sin B, cos( A + B ) − cos( A − B ) = −2 sin A sin B, , 2 sin A sin B = −[cos( A + B ) − cos( A − B )] = cos( A − B ) − cos( A + B ), , 2 sin A sin B = cos( A − B) − cos( A + B), , Or, , sin A sin B =, , 1, cos( A − B ) − cos( A − B ), 2, , 3.6.2 Transformation formulae to express product ratios into sum or difference of ratios:, consider, , sin ( A + B ) + sin ( A − B ) = 2 sin A cos B, , − − − − −(1), , sin ( A + B ) − sin ( A − B ) = 2 cos A sin B, , − − − − −(2), , cos( A + B ) + cos( A − B ) = 2 cos A cos B, , − − − − −(3), , cos( A + B ) − cos( A − B ) = −2 sin A sin B, Let, , − − − − − ( 4), , A+ B =C, , − − − − −(5), , A− B = D, , − − − − −(6), , (5) + (6), , (5) − (6), , , , A+ B+ A− B =C + D, , , , 2A = C + D, , , , A=, , , , C+D, 2, A + B − ( A − B) = C − D, , , , A+ B− A+ B =C − D, , , , 2B = C − D, , , , B=, , C−D, 2, , C+D, C−D, and B =, in equations (1),(2),(3) and (4) ,we get, 2, 2, C + D C − D, C + D C − D, sin C + sin D = 2 sin ,  cos,  sin C − sin D = 2 cos,  sin , ,  2   2 ,  2   2 , , Substituting A =, , C + D C − D, cos C + cos D = 2 cos,  cos, ,  2   2 , , C + D C − D, cos C − cos D = −2 sin ,  sin , ,  2   2 , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 56

Page 57 :
List of formulae, Transformation of product into sum or difference Transformation of sum or difference into product, C +D C −D, 1. sin A cos B = 1 sin ( A + B ) + sin ( A − B ), 1. sin C + sin D = 2 sin ,  cos, , 2,  2   2 , 2. cos A sin B = 1 sin ( A + B ) − sin ( A − B ), C +D C −D, 2, 2. sin C − sin D = 2 cos,  sin , , 2, 2, , , , , 3. cos A cos B = 1 cos( A + B ) + cos( A − B ), 2, C + D C −D, 3. cos C + cos D = 2 cos,  cos, ,  2   2 , 4. sin A sin B = 1 cos( A − B ) − cos( A + B ), 2, , C +D C −D, 4. cos C − cos D = −2 sin ,  sin , ,  2   2 , , 3.6.3 Problems on expressing sum or difference of t-ratios into product of t-ratios., 1., Express sin 5x + sin 3x in the product form., Solution:, W .K .T, C + D C − D, sin C + sin D = 2 sin ,  cos, ,  2   2 ,  5 x + 3x   5 x − 3x , sin 5 x + sin 3 x = 2 sin ,  cos, ,  2   2 ,  8x   2 x , sin 5 x + sin 3 x = 2 sin   cos  = 2 sin 4 x cos x,  2  2 , , 2., Express sin 9x + sin 2x in the product form., Solution:, W .K .T, C + D C − D, sin C + sin D = 2 sin ,  cos, ,  2   2 ,  9x + 2x   9x − 2x , sin 9 x + sin 2 x = 2 sin ,  cos, ,  2   2 ,  11x   7 x , sin 5 x + sin 3 x = 2 sin ,  cos ,  2   2 , 3., Express sin 11x − sin 5x in the product form., Solution:, W .K .T, C + D C − D, sin C − sin D = 2 cos,  sin , ,  2   2 ,  11x + 5 x   11x − 5 x , sin 11x − sin 5 x = 2 cos,  sin , , 2, 2, ,  , ,  16 x   6 x , sin 11x − sin 5 x = 2 cos,  sin   = 2 cos 8 x sin 3 x,  2   2 , 4., Express sin 7 x − sin 4x in the product form., Solution:, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 57

Page 58 :
W .K .T, C + D C − D, sin C − sin D = 2 cos,  sin , ,  2   2 ,  7x + 4x   7x − 4x , sin 7 x − sin 4 x = 2 cos,  sin , ,  2   2 ,  11x   3 x , sin 7 x − sin 4 x = 2 cos,  sin  ,  2   2 , 5., Prove that sin 6x + sin 4x = 2 sin 5x cos x ., Solution:, L.H .S = sin 6 x + sin 4 x, , W .K .T, C + D C − D, sin C + sin D = 2 sin ,  cos, ,  2   2 ,  6x + 4x   6x − 4x , L.H .S = 2 sin ,  cos, ,  2   2 ,  10 x   2 x , L.H .S = 2 sin ,  cos ,  2   2 , L.H .S = 2 sin 5 x cos x, 6., Prove that cos13x + cos 7 x = 2 cos10x cos 3x ., Solution:, L.H .S = cos 13x + cos 7 x, , C + D C − D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 ,  13x + 7 x   13x − 7 x , L.H .S = 2 cos,  cos, , 2, 2, ,  , ,  20 x   6 x , L.H .S = 2 cos,  cos  = 2 cos 10 x cos 3x = R.H .S,  2   2 , 7., Prove that cos13x − cos 7 x = −2 sin 10x sin 3x ., Solution:, L.H .S = cos 13 x − cos 7 x, , C + D C − D, W .K .T cos C − cos D = −2 sin ,  sin , ,  2   2 ,  13 x + 7 x   13 x − 7 x , L.H .S = −2 sin ,  sin , , 2, 2, ,  , ,  20 x   6 x , L.H .S = −2 sin ,  sin   = −2 sin 10 x sin 3 x = R.H .S,  2   2 , 8., Prove that cos 100 0 + cos 80 0 = 0., Solution:, L.H .S = cos1000 + cos 800, , C +D C −D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 ,  1000 + 800   1000 − 800 ,  cos, , L.H .S = 2 cos, 2, 2, ,  , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 58

Page 59 :
 180 0   20 0 ,  cos, , L.H .S = 2 cos,  2   2 , L.H .S = 2 cos 90 0 cos 10 0 = 2  0  cos 10 0 = 0 = R.H .S, 9., Prove that sin 47 0 + cos 77 0 = cos 17 0., Solution:, L.H .S = sin 47 0 + cos 77 0, , (, , L.H .S = sin 47 0 + cos 90 − 130, L.H .S = sin 47 + sin 13, 0, , 0, , ), ,  cos(90 −  ) = sin , , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 ,  47 0 + 130   47 0 − 130 ,  cos, , L.H .S = 2 sin , 2, 2, ,  , ,  60 0   34 0 ,  cos, , L.H .S = 2 sin ,  2   2 , 1, L.H .S = 2 sin 30 0 cos 17 0 = 2   cos 17 0 = cos 17 0 = R.H .S, 2, 10., Prove that cos 80 0 + cos 40 0 − cos 40 0 = 0., Solution:, L.H .S = cos 80 0 + cos 40 0 − cos 20 0, , C + D C − D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 ,  80 0 + 40 0   80 0 − 40 0 ,  cos,  − cos 20 0, L.H .S = 2 cos, 2, 2, ,  , ,  120 0   40 0 ,  cos,  − cos 20 0, L.H .S = 2 cos,  2   2 , L.H .S = 2 cos 60 0 cos 20 0 − cos 20 0 =, 1, L.H .S = 2   cos 200 − cos 200 = cos 200 − cos 200 = 0 = R.H .S, 2, 11., Prove that cos A + cos(1200 − A) + cos(1200 + A) = 0., Solution:, L.H .S = cos A + cos(1200 − A) + cos(1200 + A), , C +D C −D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 , , (, ,  1200 − A + 1200 + A   1200 − A − 1200 + A,  cos, L.H .S = cos A + 2 cos, 2, 2, ,  ,  2400   − 2 A ,  cos, L.H .S = cos A + 2 cos, ,  2   2 , , ) , , , , , L.H .S = cos A + 2 cos1200 cos A,  cos(−  ) = cos , 0, L.H .S = cos A + 2 cos(180 − 60 )cos A, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 59

Page 60 :
L.H .S = cos A + 2  − cos 600  cos A, 1, L.H .S = cos A − 2   cos A = cos A − cos A = 0 = R.H .S, 2, cos A + cos B,  A+ B, = cot, 12., Prove that, , sin A + sin B,  2 , Solution:, cos A + cos B, L.H .S =, sin A + sin B, C + D C − D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 , C + D C − D, sin C + sin D = 2 sin ,  cos, ,  2   2 ,  A+ B  A− B,  A+ B, 2 cos,  cos,  cos, , 2   2 , 2 , , , L.H .S =, =,  A+ B  A− B,  A+ B, 2 sin ,  cos,  sin , ,  2   2 ,  2 ,  A+ B, L.H .S = cot ,  = R.H .S,  2 , sin 7 A + sin 3 A, = tan 5 A cot 2 A, 13., Prove that, sin 7 A − sin 3 A, Solution:, sin 7 A + sin 3 A, L.H .S =, sin 7 A − sin 3 A, C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, sin C − sin D = 2 cos,  sin , ,  2   2 ,  7 A + 3A   7 A − 3A , 2 sin ,  cos, , 2, 2, , , ,  = 2 sin 5 A cos 2 A = tan 5 A cot 2 A = R.H .S, L.H .S =,  7 A + 3 A   7 A − 3 A  2 cos 5 A sin 2 A, 2 cos,  sin , , 2, 2, ,  , , 14., , Prove that, , sin 56 0 − sin 44 0, = cot 840., cos 56 0 + cos 44 0, , Solution:, sin 56 0 − sin 440, L.H .S =, cos 56 0 + cos 44 0, C +D C −D, W .K .T sin C − sin D = 2 cos,  sin , ,  2   2 , C +D C −D, cos C + cos D = 2 cos,  cos, ,  2   2 , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 60

Page 61 :
 56 0 + 44 0   56 0 − 44 0 ,  sin , , 2 cos, 2, 2, 2 cos 50 0 sin 6 0, , , , , L.H .S =, =,  56 0 + 44 0   56 0 − 44 0  2 cos 50 0 cos 6 0,  cos, , 2 cos, 2, 2, ,  , , , (, , ), , sin 6 0, = tan 6 0 = tan 90 − 84 0 = cot 84 0 = R.H .S, 0, cos 6, 15., Prove that sin 40 0 + sin 20 0 − cos 10 0 = 0., Solution:, L.H .S = sin 40 0 + sin 20 0 − cos 10 0, L.H .S =, , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 ,  40 0 + 20 0   40 0 − 20 0 ,  cos,  − cos 10 0, L.H .S = 2 sin , 2, 2, ,  , ,  60 0   20 0 ,  cos,  − cos 10 0, L.H .S = 2 sin , 2, 2, ,  , , 1, L.H .S = 2 sin 30 0 cos 10 0 − cos 10 0 = 2   cos 10 0 − cos 10 0, 2, 0, 0, L.H .S = cos 10 − cos 10 = 0 = R.H .S, 16., Prove that cos 40 0 + cos 80 0 + cos 160 0 = 0., Solution:, L.H .S = (cos 80 0 + cos 40 0 ) + cos 160 0, C + D C − D, W .K .T cos C + cos D = 2 cos,  cos, ,  2   2 ,  80 0 + 40 0   80 0 − 40 0 ,  cos,  + cos(180 − 20 0 ), L.H .S = 2 cos, 2, 2, ,  , ,  120 0   40 0 ,  cos,  +  − cos 20 0, L.H .S = 2 cos,  2   2 , L.H .S = 2 cos 60 0 cos 20 0 − cos 20 0, 1, L.H .S = 2   cos 20 0 − cos 20 0, 2, L.H .S = cos 20 0 − cos 20 0, L.H .S = 0 = R.H .S, , 17., , Prove that, , sin 70 0 + sin 50 0, = 3, cos 70 0 + cos 50 0, , Solution:, sin 70 0 + sin 500, L.H .S =, cos 70 0 + cos 50 0, C +D C −D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C +D C −D, cos C + cos D = 2 cos,  cos, ,  2   2 , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 61

Page 62 :
 70 0 + 50 0   70 0 − 50 0 ,  cos, , 2 sin , 2, 2, 2 sin 60 0 cos 20 0, , , , , L.H .S =, =,  70 0 + 50 0   70 0 − 50 0  2 cos 60 0 cos 20 0,  cos, , 2 cos, 2, 2, ,  , , 0, sin 60, L.H .S =, = tan 60 0 = 3 = R.H .S, 0, cos 60, , 18., , Prove that, , cos 17 0 + sin 17 0, = tan 62 0, cos 17 0 − sin 17 0, , Solution:, cos 17 0 + sin 17 0, L.H .S =, cos 17 0 − sin 17 0, cos 90 − 730 + sin 17 0, L.H .S =, cos 90 − 730 − sin 17 0, , (, (, , L.H .S =, , ), ), , sin 730 + sin 17 0, sin 730 − sin 17 0, , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, sin C − sin D = 2 cos,  sin , ,  2   2 ,  730 + 17 0   730 − 17 0 ,  cos, , 2 sin , 2, 2, ,  , , L.H .S =,  730 + 17 0   730 − 17 0 ,  sin , , 2 cos, 2, 2, ,  , , 2 sin 450 cos 280, L.H .S =, = tan 450 cot 280, 0, 0, 2 cos 45 sin 28, L.H .S = 1  cot 280 = cot 280 = cot 90 − 62 0 = tan 62 0 = R.H .S, , (, , 19., , Prove that, , Solution:, , L.H .S =, , ), , sin  + sin 3 + sin 5, = tan  ., cos  + cos 3 + cos 5, , (sin 5 + sin  ) + sin 3, sin  + sin 3 + sin 5, =, cos  + cos 3 + cos 5 (cps5 + cos  ) + cos 3, , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, cos C + cos D = 2 cos,  cos, ,  2   2 , , 20., ,  5 +    5 −  ,  6   4 , 2 sin ,  cos,  2 sin   cos ,  2   2  =,  2   2 , L.H .S =,  5 +    5 −  ,  6   4 , 2 cos,  cos,  2 cos  cos ,  2   2 ,  2   2 , 2 sin 3 cos 2 sin 3, L.H .S =, =, = tan 3 = R.H .S, 2 cos 3 cos 2 cos 3, sin x + sin 6 x + sin 11x, = tan 6 x, Prove that, cos x + cos 6 x + cos 11x, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 62

Page 63 :
Solution:, , L.H .S =, , (sin 11x + sin x ) + sin 6 x, sin x + sin 6 x + sin 11x, =, cos x + cos 6 x + cos 11x (cos 11x + cos x ) + cos 6 x, , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, cos C + cos D = 2 cos,  cos, ,  2   2 ,  11x + x   11x − x , 2 sin ,  cos, , 2   2  2 sin 6 x cos 5 x + sin 6 x, , L.H .S =, =,  11x + x   11x − x  2 cos 6 x cos 5 x + cos 6 x, 2 cos,  cos, ,  2   2 , sin 6 x(cos 5 x + 1) sin 6 x, L.H .S =, =, = tan 6 x = R.H .S, cos 6 x(cos 5 x + 1) cos 6 x, 21., , Prove that, , sin x + sin 3x + sin 5 x + sin 7 x, = tan 4 x, cos x + cos 3x + cos 5 x + cos 7 x, , Solution:, L.H .S =, , (sin 7 x + sin x ) + (sin 5 x + sin 3x ), sin x + sin 3 x + sin 5 x + sin 7 x, =, cos x + cos 3 x + cos 5 x + cos 7 x (cos 7 x + cos x ) + (cos 5 x + cos 3 x ), , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, cos C + cos D = 2 cos,  cos, ,  2   2 ,  7x + x   7x − x ,  5 x + 3x   5 x − 3x , 2 sin ,  cos,  + 2 sin ,  cos, , 2   2 , 2   2 , , , L.H .S =,  7x + x   7x − x ,  5 x + 3x   5 x − 3x , 2 cos,  cos,  + 2 cos,  cos, ,  2   2 ,  2   2 , 2 sin 4 x cos 3 x + 2 sin 4 x cos x, L.H .S =, 2 cos 4 x cos 3 x + 2 cos 4 x cos x, 2 sin 4 x(cos 3 x + cos x ) sin 4 x, L.H .S =, =, = tan 4 x = R.H .S, 2 cos 4 x(cos 3 x + cos x ) cos 4 x, sin 6 A + sin 5 A + sin 3 A + sin 2 A, = tan 4 A, 22., Prove that, cos 6 A + cos 5 A + cos 3 A + cos 2 A, Solution:, (sin 6 A + sin 2 A) + (sin 5 A + sin 3 A), sin 6 A + sin 5 A + sin 3 A + sin 2 A, L.H .S =, =, cos 6 A + cos 5 A + cos 3 A + cos 2 A (cos 6 A + cos 2 A) + (cos 5 A + cos 3 A), C + D C − D, C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos,  & cos C + cos D = 2 cos,  cos, ,  2   2 ,  2   2 ,  6A + 2A   6A − 2A ,  5A + 3A   5A − 3A , 2 sin ,  cos,  + 2 sin ,  cos, , 2, 2, 2  , 2 , , , , , , L.H .S =,  6A + 2A   6A − 2A ,  5A + 3A   5A − 3A , 2 cos,  cos,  + 2 cos,  cos, , 2, 2, 2  , 2 , ,  , , , 2 sin 4 A cos 2 A + 2 sin 4 A cos A, L.H .S ==, 2 cos 4 A cos 2 A + 2 cos 4 A cos A, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 63

Page 64 :
L.H .S =, , 23., , Prove that, , 2 sin 4 A(cos 2 A + cos A) sin 4 A, =, = tan 4 A = R.H .S, 2 cos 4 A(cos 2 A + cos A) cos 4 A, , cos x + cos 2 x − cos 3x − cos 4 x, = tan x, sin x + sin 2 x + sin 3x + sin 4 x, , Solution:, L.H .S =, , cos x + cos 2 x − cos 3 x − cos 4 x (cos x − cos 3 x ) + (cos 2 x − cos 4 x ), =, (sin 3x + sin x ) + (sin 4 x + sin 2 x ), sin x + sin 2 x + sin 3 + sin 4 x, , C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 , C + D C − D, cos C − cos D = −2 sin ,  sin , ,  2   2 , L.H .S =, , L.H .S =, L.H .S =, L.H .S =, ,  x + 3x   x − 3x ,  2x + 4x   2x − 4x , − 2 sin ,  sin ,  − 2 sin ,  sin , ,  2   2 ,  2   2 ,  3x + x   3x − x ,  4x + 2x   4x − 2x , 2 sin ,  cos,  + 2 sin ,  cos, ,  2   2 ,  2   2 , − 2 sin 2 x sin (− x ) − 2 sin 3 x sin (− x ), 2 sin 2 x cos x + 2 sin 3 x cos x, 2 sin 2 x sin x + 2 sin 3 x sin x, 2 sin 2 x cos x + 2 sin 3 x cos x, 2 sin x(sin 2 x + sin 3 x ) sin x, =, = tan x = R.H .S, 2 cos x(sin 2 x + sin 3 x ) cos x, , 3.6.3 Problems on expressing product of t-ratios into sum or difference of t-ratios., 1., Express sin 5x cos 3x into sum or difference., Solution:, 1, W .K .T sin A cos B = sin ( A + B ) + sin ( A − B ), 2, 1, sin 5 x cos 3 x = sin (5 x + 3 x ) + sin (5 x − 3 x ), 2, 1, sin 5 x cos 3 x = sin 8 x + sin 2 x , 2, 2., Express cos 8x sin 2x into sum or difference., Solution:, 1, W .K .T cos A sin B = sin ( A + B ) − sin ( A − B ), 2, 1, cos 8 x sin 2 x = sin (8 x + 2 x ) − sin (8 x − 2 x ), 2, 1, cos 8 x sin 2 x = sin 10 x − sin 6 x , 2, 3., Express 2 cos 8x cos 6x into sum or difference., Solution:, W .K .T 2 cos A cos B = cos( A + B ) + cos( A − B ), 2 cos 8 x cos 6 x = cos(8 x + 6 x ) + cos(8 x − 6 x ), 2 cos 8 x cos 6 x = cos 14 x + cos 2 x, Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 64

Page 65 :
4., Express 4 sin 8x sin 5x into sum or difference., Solution:, W .K .T 2 sin A sin B = cos( A − B ) − cos( A + B ), 4 sin 8 x sin 5 x = 2(2 sin 8 x sin 5 x ), = 2{cos (8 x − 5 x ) − cos(8 x + 5 x )}, = 2{cos 3 x − cos 14 x}, 5., Express sin 1250 sin 550 into sum or difference., Solution:, W .K .T 2 sin A sin B = cos( A − B ) − cos( A + B ), 1, sin 1250 sin 550 = cos(1250 − 550 ) − cos(1250 + 550 ), 2, 1, = cos 70 0 − cos 180 0, 2, 1, = cos 70 0 − cos(90 + 90 0 ), 2, 1, = cos 70 0 −  − cos 90 0, 2, 1, = cos 70 0 + 1, 2, 1, 6., Show that cos 20 0 cos 40 0 cos 60 0 cos 80 0 = ., 16, Solution:, L.H .S = cos 20 0 cos 40 0 cos 60 0 cos 80 0, , , , , , , , , , , , , , , , , , , , , , (, , ), , = cos 60 0 cos 40 0 cos 20 0 cos 80 0, W .K .T cos A cos B =, ,  (, , 1, cos( A + B ) + cos( A − B ) & cos 600 = 1, 2, 2, , ), , (, , ), , 1 1,  cos 40 0 + 20 0 + cos 40 0 − 20 0 cos 80 0, 2 2, 1, = cos 60 0 + cos 20 0 cos 80 0, 4, 11, , =  + cos 20 0  cos 80 0, 42, , 1, 1, = cos 80 0 + cos 80 0 cos 20 0, 8, 4, 1, 1 1, = cos 180 − 100 0 +  cos 80 0 + 20 0 + cos 80 0 − 20 0, 8, 4 2, 1, 1, =  − cos 100 0 + cos 100 0 + cos 60 0, 8, 8, 1, 1, 1, = − cos 100 0 + cos 100 0 + cos 60 0, 8, 8, 8, 1 1 1, =  =, = R.H .S, 8 2 16, 1, 7., Show that cos 20 0 cos 40 0 cos 80 0 = ., 8, Solution:, L.H .S =, , (, , ), , (, ,  (, , ), , (, , ), , (, , ), , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 65

Page 66 :
L.H .S = cos 20 0 cos 40 0 cos 80 0, , (, , ), , = cos 40 0 cos 20 0 cos 80 0, W .K .T cos A cos B =, ,  (, , 1, cos( A + B ) + cos( A − B ), 2, , ), , ), , (, , 1, cos 40 0 + 20 0 + cos 40 0 − 20 0 cos 80 0, 2, 1, = cos 60 0 + cos 20 0 cos 80 0, 2, 11, , =  + cos 20 0  cos 80 0, 22, , 1, 1, = cos 80 0 + cos 80 0 cos 20 0, 4, 2, 1, 1 1, = cos 180 − 100 0 +  cos 80 0 + 20 0 + cos 80 0 − 20 0, 4, 2 2, 1, 1, =  − cos 100 0 + cos 100 0 + cos 60 0, 4, 4, 1, 1, 1, = − cos 100 0 + cos 100 0 + cos 60 0, 4, 4, 4, 1 1 1, =  = = R.H .S, 4 2 8, 3, ., 8., Show that sin 20 0 sin 40 0 sin 80 0 =, 8, Solution:, L.H .S = sin 20 0 sin 40 0 sin 80 0, L.H .S =, , (, , ), , (, ,  (, , ), , ), , (, , (, , (, , ), , ), , ), , = sin 40 0 sin 20 0 sin 80 0, W .K .T sin A sin B =, ,  (, , 1, cos( A − B ) − cos( A + B ), 2, , ), , ), , (, , 1, cos 40 0 − 20 0 − cos 40 0 + 20 0 sin 80 0, 2, 1, 1, 1, = cos 20 0 − cos 60 0 sin 80 0 =  cos 20 0 −  sin 80 0, 2, 2, 2, 1, 1, = sin 80 0 cos 20 0 − sin 80 0, 2, 4, 1, W .K .T sin A cos B = sin ( A + B ) + sin ( A − B ), 2, 1 1, 1, L.H .S =  sin 80 0 + 20 0 + sin 80 0 − 20 0 − sin 180 − 100 0, 2 2, 4, 1, 1, = sin 100 0 + sin 60 0 − sin 100 0, 4, 4, 1, 1, 1, = sin 100 0 + sin 60 0 − sin 100 0, 4, 4, 4, 1, 1, 3, 3, L.H .S = sin 60 0 = , =, = R.H .S, 4, 4 2, 8, L.H .S =, , , , , ,  (, , (, , ), , (, , ), , (, , ), , ), , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 66

Page 67 :
9., , Show that sin 20 0 sin 40 0 sin 60 0 sin 80 0 =, , 3, ., 16, , Solution:, L.H .S = sin 20 0 sin 40 0 sin 60 0 sin 80 0, , (, , ), , = sin 60 0 sin 40 0 sin 20 0 sin 80 0, 1, cos( A − B ) − cos( A + B ) & sin 60 0 = 3, 2, 2, , W .K .T sin A sin B =, ,  (, , ), , ), , (, , 3 1,  cos 40 0 − 20 0 − cos 40 0 + 20 0 sin 80 0, 2 2, 3, 3, 1, 0, 0, =, cos 20 0 − cos 60 0 sin 80 0 =,  cos 20 −  sin 80, 4, 4 , 2, , L.H .S =, , , , , , 3, 3, sin 80 0 cos 20 0 −, sin 80 0, 4, 8, 1, W .K .T sin A cos B = sin ( A + B ) + sin ( A − B ), 2, 3 1, 3, L.H .S =,  sin 80 0 + 20 0 + sin 80 0 − 20 0 −, sin 180 − 100 0, 4 2, 8, 3, 3, =, sin 100 0 + sin 60 0 −, sin 100 0, 8, 8, 3, 3, 3, =, sin 100 0 +, sin 60 0 −, sin 100 0, 8, 8, 8, 3, 3, 3 3, L.H .S =, sin 60 0 =, , =, = R.H .S, 8, 8, 2 16, 1, 10., Show that sin 10 0 sin 50 0 sin 70 0 = ., 8, Solution:, L.H .S = sin 100 sin 500 sin 700, =, ,  (, , ), , (, , (, , ), , (, , (, , ), , ), , ), , = sin 500 sin 10 0 sin 700, W .K .T sin A sin B =, ,  (, , 1, cos( A − B ) − cos( A + B ), 2, , ), , (, , ), , 1, cos 50 0 − 10 0 − cos 50 0 + 10 0 sin 70 0, 2, 1, 1, 1, = cos 40 0 − cos 60 0 sin 70 0 =  cos 40 0 −  sin 70 0, 2, 2, 2, 1, 1, = sin 70 0 cos 40 0 − sin 700, 2, 4, 1, W .K .T sin A cos B = sin ( A + B ) + sin ( A − B ), 2, 1 1, 1, L.H .S =  sin (70 0 + 400 ) + sin (700 − 400 ) − sin (180 − 110 0 ), 2 2, 4, 1, 1, = (sin 110 0 + sin 300 ) − sin 110 0, 4, 4, L.H .S =, , , , , , , , , , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 67

Page 68 :
1, 1, 1, sin 110 0 + sin 30 0 − sin 110 0, 4, 4, 4, 1, 1 1 1, L.H .S = sin 30 0 =  = = R.H .S, 4, 4 2 8, =, , 3.6.4 Problems on proving conditional identities, 1., If A + B + C = 1800 then prove that sin 2 A + sin 2B + sin 2C = 4 sin A sin B sin C, Solution:, L.H .S = sin 2 A + (sin 2 B + sin 2C ), C + D C − D, W .K .T sin C + sin D = 2 sin ,  cos, ,  2   2 ,  2 B + 2C   2 B − 2C , L.H .S = sin 2 A + 2 sin ,  cos, , 2, 2, ,  , , L.H .S = sin 2 A + 2 sin (B + C ) cos(B − C ), Use, , sin 2 A = 2 sin A cos A & B + C = 180 0 − A, , (, , ), , L.H .S = 2 sin A cos A + 2 sin 180 0 − A cos(B − C ), L.H .S = 2 sin A cos A + 2 sin A cos(B − C ), L.H .S = 2 sin Acos A + cos(B − C ), If, , A + B + C = 180 0  A = 180 0 − (B + C ), ,  , , , , , , L.H .S = 2 sin A cos 180 0 − (B + C ) + cos(B − C ), L.H .S = 2 sin A− cos(B + C ) + cos(B − C ), L.H .S = 2 sin Acos(B − C ) − cos(B + C ), , Use, , C + D C − D, cos C − cos D = −2 sin ,  sin , ,  2   2 , , ,  B − C − ( B + C )   B − C + B + C , L.H .S = 2 sin A− 2 sin ,  sin , , 2, 2, ,  , , ,  − 2C   2 B , L.H .S = −4 sin A sin ,  sin , ,  2   2 , L.H .S = −4 sin A sin( −C ) sin B, L.H .S = 4 sin A sin C sin B, ,  sin (−  ) = − sin , , L.H .S = 4 sin A sin B sin C = R.H .S, 2., If A + B + C = 1800 then prove that sin 2 A − sin 2B + sin 2C = 4 cos A cos C sin B, Solution:, L.H .S = (sin 2 A − sin 2 B ) + sin 2C, C +D C −D, W .K .T sin C − sin D = 2 cos,  sin , ,  2   2 ,  2 A + 2B   2 A − 2B , L.H .S = 2 cos,  sin ,  + sin 2C, 2, 2, ,  , , = 2 cos( A + B ) sin( A − B ) + sin 2C, Use, , sin 2C = 2 sin C cos C & A + B + C = 1800  A + B = 1800 − C, , L.H .S = 2 cos(1800 − C ) sin( A − B) + 2 sin C cos C, L.H .S = 2  − cos C sin( A − B) + 2 sin C cos C, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 68

Page 69 :
L.H .S = 2 cos C − sin( A − B) + sin C , , , , , , L.H .S = 2 cos C − sin( A − B) + sin 180 − ( A + B), L.H .S = 2 cos C − sin( A − B) + sin( A + B), , , ,  C = 180 0 − ( A + B), , L.H .S = 2 cos C sin( A + B) − sin( A − B), , L.H .S = 2 cos C sin( A + B) − sin( A − B), C + D, W .K .T sin C − sin D = 2 cos,  sin D,  2 ,  A + B + A − B   A + B − (A − B) , L.H .S = 2 cos C  2 cos,  sin , , 2, 2, ,  , ,  2 A   2B , L.H .S = 4 cos C cos,  sin , ,  2   2 , L.H .S = 4 cos C cos A sin B = 4 cos A cos C sin B, , Engineering Mathematics_Unit-3, Trigonometry_Ramachandra Sutar, , Page 69