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Trigonometry, 1) The unit of angle measure is degree and it is denoted by 0., 2) The unit of angle measure is radian and it is denoted by c., 3) π radian =180 degree i.e.  C  1800 ,, 0, , , , , 1radian =  180  degree i.e. 1C   180  ,, , , 4)1degree =, , , , , , , , , , ,  180 , , , radian i.e.10   , , , , , C, , ,  ,,  180 , , Trigonometric Ratios, The ratios which relate the sides of a right angle to its angles are called trigonometric ratios., There are six trigonometric ratios of an angle  . They are sin  , cos  , tan  , cot  ,sec  and cosec, Consider ABC is a right angled triangle as shown in the figure where A   . AB is, adjacent side, BC is opposite side and AC is hypotenuse, then, , Oppositeside BC, Adjacent side AB, , ,, 2) cosθ =, , ,, Hypotenuse, AC, Hypotenuse, AC, Hypotenuse, AC, Hypotenuse, AC, 4) cosecθ =, , , 5)secθ =, , ,, Oppositeside BC, Adjacent side AB, , 1)sinθ =, , Oppositeside BC, , ,, Adjacent side AB, Adjacent side AB, 6) cot θ =, , ., Oppositeside BC, , 3) tanθ =

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Trigonometric functions of standard angles, Angle, , 00, , 300, , 450, , 600, , 900, , Ratio, sin, ,  ( / 4)c, ,  ( / 3)c, , 0, ,  ( / 6)c, 1, 2, , 1, 2, , 3, 2, ,  ( / 2)c, 1, , cos, , 1, , 3, 2, , 1, 2, , 1, 2, , tan, , 0, , 1, 3, , 1, , cosec, , , , 2, , 2, , sec, , 1, , 2, 3, , 2, , cot, , , , 1, , 3, , 1800, , , 2700, , 3600, , 0, ,  (3 / 2)c, -1, ,  (2 )c, 0, , 0, , -1, , 0, , 1, , , , 0, , , , 0, , 2, 3, , 1, , , , -1, , , , 2, , , , -1, , , , 1, , 1, 3, , 0, , , , 0, , , , 3, , Signs of trigonometric functions, Quadrants, , Trigonometric, ratios, sin, cos, tan, , I, , II, , III, , IV, , +, +, +, , +, -, , +, , +, -, , From above table, we say that, In I quadrant – all are positive., In II quadrant – sine and cosec are positive, In III quadrant – tan and cot are positive, In IV quadrant – cos and sec are positive.

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Trigonometric Ratios of Compound and Allied Angles, Compound Angle [BTE 2016] – An angle obtained by algebraic sum or difference of two or more, angles is called a compound angle., e.g. If A,B,C--- are angles, then A  B, A  B, A  B  C , A  B  C    are compound angles., Allied Angles – If the sum or difference of the measures of two angles is either zero or is an integral, , multiple of 900 or n where n  I , then these angles are called allied angles., 2, e.g. If is the measure of a given angle, then its allied angles are of the form 0,, , 3,  ,,   , 2   , 2   etc., 2, 2, Trigonometric functions of addition and subtraction, 1) sin( A  B)  sin Acos B  sin B cos A,, 2) sin( A  B)  sin Acos B  sin B cos A,, , 3) cos( A  B)  cos Acos B  sin Asin B,, 5)tan( A  B)  tan A  tan B ,, 1  tan A.tanB, , 4) cos( A  B)  cos Acos B  sin Asin B,, 6)tan( A  B)  tan A  tan B ,, 1  tan A.tanB, , 7)sin( A  B)sin( A  B)  sin 2 A  sin 2 B,, , 8)cos( A  B)cos( A  B)  cos 2 A  cos 2 B., , Trigonometric functions of allied angles, , , Angle, Ratio, sin, cos, tan, cosec, sec, cot, , , 2, ,  sin , cosθ,  tan , cosecθ, secθ, cotθ, , , , cosθ, sin , cotθ, secθ, cosecθ, tan , , , 2, , , , cosθ,  sin , cotθ, secθ, cosecθ,  tan , ,  , ,  , , 3, , 2, , 3, , 2, , 2 , , 2 , , sin , cosθ,  tan , cosecθ, secθ, cotθ, ,  sin , cosθ, tan , cosecθ, secθ, cotθ, , cosθ,  sin , cotθ, secθ, cosecθ, tan , , cosθ, sin , cotθ, secθ, cosecθ,  tan , ,  sin , cosθ,  tan , cosecθ, secθ, cotθ, , sin , cosθ, tan , cosecθ, secθ, tan , , Examples:, Without using the calculator, find the value of, , , 3  1, 3  1, 0, 0, 1) cos(75)0 . [BTE 2017] cos(75)0 ,  2) tan(75) [BTE2012]  tan(75) , , 2 2 , 3  1, , , 3) sin2100 [ -1/2], , 4) sec36600 [2], , 5) sin1500  cos3000  tan 3150  sec2 36600 [6], , [sin1500  1/ 2, cos3000  1/ 2, tan 3150  1,sec36600  2], 6), , sec2 1350, cos(2400 )  2sin 9300, , [4]

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Examples for Tutorial, Without using the calculator, prove that, 1) sin 4200 cos3900  cos(3000 )sin(3300 )  1 [BTE2016], , [sin 4200  3 / 2, cos3900  3 / 2, cos(3000 )  1/ 2, sin(3300 )  1/ 2], 2) sin1500  tan 3150  cos3000  sec2 3600  3 [BTE2015], 0, 0, 0, 0, , [sin150  1/ 2, tan 315  1, cos300  1/ 2, sec360  1], , ----------------------------------------------------------------------------------------------------------------------------------, , Examples:, Prove that, 1) sin(n  1) A.sin(n  2) A  cos(n  1) A.cos(n  2) A  cos A, 2), , sin( A  B) cot A  cot B, , cos( A  B) 1  cot A cot B, , 5, , 1, and tan y  , prove that x  y  . [BTE2014], 6, 4, 11, 4) In any ABC , prove that tan A  tan B  tan C  tan A.tan B.tan C . [BTE2014], , 3) If tan x , , [ A  B  1800  C, tan(1800  C)   tan C], 1  tan 2 .tan  cos 3, , 5) Prove that, . [BTE2013], 1  tan 2 .tan , cos , 4 , 5 3,  B  2 , find a) sin( A  B) , b) cos( A  B) ., 6) If sin A  ,  A   and cos B  ,, 5 2, 13, 56, 63, [sin( A  B)  , cos( A  B)   ], 65, 65, , 2, , Examples for Tutorial, 0, 0, 0, 0, 1) sin(45  A).cos(45  B)  cos(45  A).sin(45  B)  cos( A  B), , 2), , sin( A  B) sin( B  C ) sin(C  A), , , 0, sin A sin B sin B sin C sin C sin A, , sin( A  B) tan A  tan B, , sin( A  B) tan A  tan B, 1, 1, 4) If tan A  and tan B  , find tan( A  B) .[BTE2016] [tan( A  B)  1], 2, 3, 5) Show that tan3A  tan 2 A  tan A  tan3A.tan 2 A.tan A . [3 A  2 A  A], 6) Prove that 1  tan  .tan 2  sec2 . [BTE2015], 5, 7, 7) If sin    , cos   , and  ,  lies in the third quadrant, find sin(   ) . [BTE2013], 13, 25, , 3), , 12, 24, , , cos    13 ( is in III quadrant, cos  is negative.), sin    25 (  is in III quadrant, sin  is negative.) , , , sin(   )   253, , , , 325

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Trigonometric Ratios of Multiple and Sub-multiple Angles, Multiple Angles – Angles of the form 2 ,3 , 4    are integral multiples of  . They are called, multiple angles.,  3, Sub-multiple Angles – Angles of the form , ,    are called sub-multiple angles of  ., 2 2, Trigonometric functions of double and triple angles:, , 1) sin 2  2sin  cos  ,, , 2) cos 2  cos 2   sin 2   1  2sin 2   2 cos 2   1,, , 3) sin 2  2 tan  ,, 1  tan 2 , , 2, 4) cos 2  1  tan  ,, 5) tan 2  2 tan , 2, 1  tan , 1  tan 2 , 6)1  cos 2  2sin 2  ,  sin 2   1  cos 2 ,, 7)1  cos 2  2 cos 2  ,, 2, , 8) 1  sin 2  (cos   sin  )2 ,, 10) sin 3  3sin   4sin 3  ,, ,  cos 2   1  cos 2 ,, 2, , 9) 1  sin 2  (cos   sin  ) 2 ,, 11) cos 3  4 cos3   3cos  ,, , 3, 12) tan 3  3 tan   tan  ., 1  3 tan 3 , , Trigonometric functions of half angles:, 1) sin   2sin( / 2) cos( / 2),, 3) sin  , , 2 tan( / 2), 1  tan ( / 2), 2, , 2) cos   cos 2 ( / 2)  sin 2 ( / 2)  1  2sin 2 ( / 2)  2 cos 2 ( / 2)  1,, 4) cos  , , ,, , 6)1  cos   2sin 2 ( / 2),, , 1  tan 2 ( / 2), 1  tan ( / 2), 2, , ,, , 5) tan  , , 2 tan( / 2), 1  tan 2 ( / 2), , 7)1  cos   2 cos 2 ( / 2),, , 8)1  sin    cos( / 2)  sin( / 2)  ,, , 9)1  sin   cos( / 2)  sin( / 2) , , 2, , 2, , ., , Examples, 1) If   450 , verify that, b) tan 3 , , a) sin 3  3sin   4sin3 , , 2) If cos   , 3) If sin A , , 3tan   tan3 , , 12, 3, and    , , find a) sin 2, 13, 2, , 1  3tan 2 , 120 , b) cos 2, 169 , , 1, , find sin 3A . [sin 3 A  1] [BTE2014], 2, , 4) Prove that, a), , c), , sin   sin 2,  tan , 1  cos   cos 2, , b), , 2  2  2  2 cos8  2 cos , , sin 9 cos 9, , 2, sin 3 cos 3, , 119 , 169 

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Examples for Tutorial, , 1) If   600 , verify that, a) sin 2  2sin  cos, , b) cos 2  2cos2   1, , 2) If sin   0.6 , find sin 3 . [sin 3  0.936], 3) If sin   0.4 , find cos3 . [cos 3  0.3297], 4) If sin A  0.4 , find sin 3A . [sin 3 A  0.944] [BTE2014], sec 4  1 tan 4, sin 2  cos , ,  cot , 5) Prove that a), b), 1  sin   cos 2, sec 2  1 tan , c), , 2  2  2 cos 4  2 cos , , Examples on half angle formulae, 1) If   450 , find the values of sin( / 2) and cos( / 2) without using the calculator., , [Use the formulae 1  cos   2sin 2 ( / 2), 1  cos   2 cos 2 ( / 2), sin( / 2) , , 2 2, 2 2, , cos( / 2) , ], 2, 2, , 2) Prove that a), , 1  sin A, sin( / 2)  sin ,  0 ,  tan( / 2), b) tan  45   , 2, 1  sin A, 1  cos( / 2)  cos , , Examples for Tutorial, , 1) If tan( / 2)  3 , find cos. [cos   cos120  1/ 2], 1, , find sin A [sin A  3 / 2], 3, 1  sin   cos ,  tan( / 2) b), 3) Prove that a), 1  sin   cos , , 2) If tan( A / 2) 

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Factorization and Defactorization Formulae, Factorization – The process of conversion from sum/difference into product is called factorization., Defactorization - The process of conversion from product of terms into sum/difference is called, defactorization., Factorization Formulae:, 1) sin C  sin D  2sin C  D cos C  D , 2) sin C  sin D  2 cos C  D sin C  D ,, 2, 2, 2, 2, 3) cos C  cos D  2 cos C  D cos C  D ,, 2, 2, , , , , , , ,  ,  , , , ,  , , , , , ,  , , , , , ,  , , , , , , 4) cos C  cos D  2sin C  D sin D  C  2sin C  D sin C  D ., 2, 2, 2, 2, , Defactorization Formulae:, , 1) 2sinA cos B  sin( A  B)  sin( A  B),, 3) 2cosAcosB  cos( A  B)  cos( A  B),, , 2) 2cosA sin B  sin( A  B)  sin( A  B),, 4) 2sinA sin B  cos( A  B)  cos( A  B)., , Examples, A) Express the following as a sum or difference of trigonometric functions., 1) 2sin 4 cos 2 [2sin 4 cos 2  sin 6  sin 2 ], 2) 2cos1170 sin 530 [2cos1170 sin 530  sin1700  sin 640 ], B) Express the following as product of trigonometric functions:, 1) sin 2  sin 4 [2sin 3 cos  ], 2) cos 700  cos 200 [2cos 450 cos 250 ], C) 1) If 2cos 600.cos100  cos A  cos B , then find A and B. [BTE2013] [ A  70, B  50], cos 3 A  2 cos 5 A  cos 7 A, sin 4  sin 2,  cos 2 A  sin 2 A.tan 3 A,  tan  b), 2) Prove that a), cos 4  cos 2, cos A  2 cos 3 A  cos 5 A, sin190  cos110, c),  3 sin190  cos(900  190 )  cos 710 , cos190  sin(900  190 )  sin 710 , 0, 0, , , cos19  sin11, d) cos100 cos500 cos 700 , , 3, 8, , e) sin100 sin 300 sin 500 sin 700 , , 2, 2,  A B , f)  cos A  cos B    sin A  sin B   4.cos2 , ,  2 , , Examples for Tutorial, A) Express the following as a sum or difference of trigonometric functions., [2 cos 4 cos 2  cos 6  cos 2 ], 1) 2cos 4 cos 2, , , 1, 2) sin( / 4) sin(3 / 4), sin( /4)sin(3 /4)  cos( /2)cos  , 2, , , , 1, 16

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B) Express the following as product of trigonometric functions:, 1) sin 7  sin5 [2 cos 6 sin  ], , 2) cos, , , 13, ,  cos, , 3,  , 2 , 2sin, sin , , 26, 26 , 13 , , C) 1) If sin800  sin 500  2sin .cos  , find  and  . [  650 ,   150 ], sin A  2sin 2 A  sin 3 A, cos 2 B  cos 2 A,  tan 2 A,  tan( A  B), b), cos A  2 cos 2 A  cos 3 A, sin 2 A  sin 2 B, 3, cos110  sin110, c), d) sin 200 sin 400 sin 600 sin 800 , [BTE2016],  tan 560, 0, 0, 16, cos11  sin11, , 2) Prove that a), , e) 8sin 200 sin 400 cos100  3, , 2, 2,  A B , f)  cos A  cos B    sin A  sin B   4.sin 2 , ,  2 , , Inverse Trigonometric Functions, Definition - If sin   x , then   sin 1 x is inverse trigonometric function. It is read as sine inverse, of x ., Similarly cos1 x, tan 1 x,cot 1 x,sec1 x,cosec1x are inverse trigonometric functions., Relation between inverse trigonometric ratios :, 1) sin 1 ( x)  cosec1 (1/ x),, cosec1 ( x)  sin 1 (1/ x),, cos 1 ( x)  sec1 (1/ x),, , sec 1 ( x)  cos 1(1/ x),, , tan 1 ( x)  cot 1 (1/ x),, , cot 1 ( x)  tan 1 (1/ x)., , 2) sin 1 ( x)   sin 1 ( x),, , cos 1 ( x)    cos 1 ( x),, , tan 1 ( x)   tan 1 ( x),, , cot 1 ( x)    cot 1 ( x),, , cosec1 ( x)   cosec1 ( x),, , sec1 ( x)    sec1 ( x)., , 3) sin 1 (sin x)  x,, cot 1 (cot x)  x,, , cos 1 (cos x)  x,, , tan 1(tan x)  x,, , cosec1 (cosecx)  x,, , sec 1(sec x)  x., , 4) sin 1 x  cos 1 x   / 2,, , tan 1 x  cot 1 x   / 2,, , cosec1 x  sec1 x   / 2., ,  x y , 5) If x  0, y  0 and xy  1then, tan 1 x  tan 1 y  tan 1 , .,  1  xy ,  x y , 6) If x  0, y  0 and xy  1then, tan 1 x  tan 1 y    tan 1 , .,  1  xy ,  x y , 7) If x  0, y  0 then, tan 1 x  tan 1 y  tan 1 , .,  1  xy , 1, Note – 1) sin 1 x , sin x

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Examples : Prove that, 1  1 , 1  1 , 1  9 , 1) tan    tan    cot   [BTE2017], 7,  13 , 2, 6, 1, 2) cot 1    tan 1    sec1 2 . [BTE 2017], 5,  11 , 3,  8,  77 , 3) sin 1    sin 1    sin 1   [BTE 2015, 2016], 5,  17 ,  85 , 4, 3,  27 , 4) cos 1    tan 1    tan 1   [BTE 2015,17], 5, 5,  11 , ,  , , Examples for Tutorial, 3, 1,  8, 1 ,  84 , 1) tan 1    tan 1    [BTE 2017], 2) sin 1    sin 1    cos 1   [BTE 2017], 5, 2,  17 , 3 4,  85 , 4,  12 ,  33 , 3) cos 1    cos 1    cos 1   [BTE 2016], 5,  13 ,  65 , 4,  12 ,  63 , 4) cos 1    cos 1    cos 1   [BTE 2017], 5,  13 ,  65 , 1  12 , 1  3 , 1  56 , 5) cos    sin    sin   . [BTE 2017],  13 , 5,  65 , 1  3 , 1  8 , 1  84 , 6) sin    sin    cos   . [BTE 2017], 5,  17 ,  85 , , Principal Value of Inverse Function, Definition – The smallest numerical value, either positive or negative, of an inverse trigonometric, function is called as principal value of the function., 1, 1, 1, e.g. We know sin 300  ,sin1500  ,sin 3900    , 2, 2, 2, 1, 1, 1,  300  sin 1   , 1500  sin 1   , 3900  sin 1     , 2, 2, 2, 1,  sin 1   have many values 300 ,1500 , 3900 ,   , 2, 1,  Principal value of sin 1   is 300., 2

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Examples: Find the principal value of the following:, ,  1 , 1) sin 1 , ,  2, ,  ,  , 4, , ,  3 , 2) sec cos1 ,  2  , , ,  , , ,  1 , 3) cos   sin 1     [BTE2012],  2 , 2, ,  2 , , ,  3, ,  1,  ,  2, , Examples for Tutorial, ,  1   2 , 1) cos1  , ,  , 2  3 , , , , , 2) tan 1  , , , , 1   ,   , 3  6, ,  , ,  4, , 3) tan 1  1  , , ---------------------------------------------------------------------------------Syllabus for Mid Semester Examination - Oct. 2018, Unit I – Logarithm, Partial Fraction, Complex Numbers, Unit II – Determinant, Matrices