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Trigonometric function of Multiple and Submultiple angles, Double Angle Identity, Submultiple Angle Identity, sine, function, sin 2A = 2 sin A cos A, sin A = 2 sin A/2cos A/2, 2 tan A, sin 2A =, 2 tan A/2, %3D, 1+tan2A, sin A =, 1+tan-A/2, cosine, cos 2A = cos? A – sin² A, cos A = cos? A/2 – sin² A/2, function, cos 2A = 1 – 2 sin² A, cos A = 1-2 sin? A/2, cos 2A = 2 cos? A -1, cos A = 2 cos A/2 - 1, cos 2A = 1- tan A, 1+tan2A, 1- tan A/2, 1+tan A/2, cos A =, 2 tan A/2, tangent, function, 2 tan A, tan 2A, tan A =, %3D, 1-tan2A, 1-tan A/2, 1-cos 2A, sin'A, 2 sin? A =1-cos 2A, 2, 2 cos²A =1+cos 2A, 1+cos 2A, cos A =

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Part A, Example 1:, If sin A == find the value of cos 2A, Solution:, cos 2A = 1- 2 sin²A, 1- 2 x-, 4, 1, 2, Example 2:, Simplify : (i) 2 sin 30° cos 30°, 2 tan 300, (ii), 1-tan2300, Solution:, (i) 2 sin 30° cos 30°, sin 2 (30°), = sin 60°, V3, 2, 2 tan 30°, (ii), tan 2 (30°), 1-tan? 30°, tan 60°, V3, 112

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Example 3:, sin A, Prove that, tan A/2, 1+ cos A, Solution:, sin A, 2 sin A/2 cos A/2, LHS,, 1+ cos A, 2 cos? A/2, sin A/2, coS A/2, tan A/2, Part B & C, 1), Prove that, sin 2 A, 1°, = cot A. Hence deduce the value of cot 22-, 1-cos 2 A, Solution:, sin 2 A, 2 sin A cos A, LHS, 1-cos 2 A, 2 sin?A, cos A, sin A, = cot A =RHS, sin 2A, We have, cot A, 1- cos 2A, Put A = 22., sin 2 (22늘), 1- cos 2 (22"), cot 22 1°, sin 45°, 1-cos 450, 1-, V2 +1, %3D, V2, 1, ||

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2), Prove that, 1- cos A+ sin A, = tan A/2, 1+ cos A+ sin A, Solution:, A, cos, А, 2 sin?+ 2 sin, 1- cos A+ sin A, LHS,, 1+ cos A+ sin A, 2 cos?, + 2 sin, cos, 2., 2., 2 sin, 2, A, A, sin-, + cos, 2, 2 cos, A, A, cos, + sin, = RHS, tan, 2, 1, 3), If tan a = -, tan B:, =- show that 2 a + B, 4, Solution:, tan 2a + tan B, tan (2 α +β)-, 1- tan 2a tan B, 2 tana, Now,, tan 2 a, %3D, 1- tan2 a, 2 G, %3D, 1-4), 3, tan 2 a, %3D, 4, lan (2 a + B)=, 21+4, 28, 3, 1, 28, 25/28, 1, 25/28, : 2 a + B, %3D, 4, Hence proved., A12A2

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EXERCISE, 1) If cos A = find the value of sin 2A., 2) Find the value of tan 2A if tan A, 2, 2 tan 15°, 3) Simplify : (i), 1+tan? 150, (ii) cos?15° - sin?15°, (iii) 2 sin 75° cos 75°, 1+sin 2A – cos 2A, 4) Prove that, = tan A, 1+sin 2A + cos 2A, sin ZA, 5) Prove that, = tan A and hence deduce the value of tan 22=, 1+cos 2A, 2, sin 4+ sin A, 6) Prove that, A, = tan, 1+ cos, + cos A, Solution:, 2 tan15°, 2 tan A, 3) i), = sin 2(15°), sin 2A, 1+ tan 15°, 1+ tan A, 1, = sin 30°, %3D, ii), :os'15° – sin?15° = c, = cos (2 x 15°), cos²A- sin2A, = cos 2A, V3, 2, = cos 30°, %3D, ii), ! sin 75° cos 75° = sin (2 x A), sin 2A = 2 sin A cos A, = sin (2 × 75°), = sin 150° = sin (180 – 30), 1, =+ sin 30°, 2