Prove the following: sin 2 x + sin 2 y sin 2 x - sin 2 y = tan ( x + y ) tan ( x - y )

LHS = sin 2 x + sin 2 y sin 2 x - sin 2 y We know that, sin A + sin B = 2 sin A + B 2 cos A - B 2 and sin A - sin B = 2 cos A + B 2 sin A - B 2 . ⇒ LHS = 2 cos 2 x + 2 y 2 cos 2 x - 2 y 2 2 cos 2 x + 2 y 2 sin 2 x - 2 y 2 = sin ( x + y ) · cos ( x - y ) cos ( x + y ) · sin ( x - y ) = tan ( x + y ) · cot ( x - y ) = tan ( x + y ) tan ( x - y ) = RHS Hence, proved.

Prove the following: sin x - sin 3 x + sin 5 x - sin 7 x cos x - cos 3 x - cos 5 x + cos 7 x = cot 2 x

LHS = sin x - sin 3 x + sin 5 x - sin 7 x cos x - cos 3 x - cos 5 x + cos 7 x = sin x - sin 7 x + sin 5 x - sin 3 x cos x + cos 7 x - cos 5 x + cos 3 x We know that, sin A - sin B = 2 cos A + B 2 sin A - B 2 and cos A + cos B = 2 cos A + B 2 cos A - B 2 = 2 cos x + 7 x 2 sin x - 7 x 2 + 2 cos 5 x + 3 x 2 cos 5 x - 3 x 2 2 cos x + 7 x 2 cos x - 7 x 2 - 2 cos 5 x + 3 x 2 cos 5 x - 3 x 2 = 2 cos 4 x sin ( - 3 x ) + 2 cos 4 x sin x 2 cos 4 x cos ( - 3 x ) - 2 cos 4 x cos x = 2 cos 4 x - sin ( 3 x ) + sin x 2 cos 4 x cos 3 x - sin x = sin x - sin 3 x cos 3 x - cos x = 2 cos 2 x sin ( - x ) - 2 sin ( x ) sin ( 2 x ) ∵ sin A - sin B = 2 cos A + B 2 sin A - B 2 & cos A - cos B = - 2 sin A - B 2 sin A + B 2 = cos 2 x sin ( 2 x ) = cot 2 x = RHS Hence, proved.

Prove the following: cos 20 ° cos 40 ° cos 60 ° cos 80 ° = 1 16

LHS = cos 20 ° cos 40 ° cos 60 ° cos 80 ° = 1 2 2 cos 20 ° cos 40 ° × 1 2 × cos 80 ° ∵ cos 60 ° = 1 2 = 1 4 cos ( 20 ° + 40 ° ) + cos ( 20 ° - 40 ° ) cos 80 ° ∵ 2 cos A cos B = cos ( A + B ) + cos ( A - B ) = 1 4 cos 60 ° + cos 20 ° cos 80 ° = 1 4 cos 60 ° cos 80 ° + cos 20 ° cos 80 ° = 1 4 1 2 cos 80 ° + 1 4 cos 20 ° cos 80 ° = 1 8 cos 80 ° + 1 4 1 2 2 cos 20 ° cos 80 ° = 1 8 cos 80 ° + 1 8 cos ( 20 ° + 80 ° ) + cos ( 20 ° - 80 ° ) = 1 8 cos 80 ° + 1 8 cos 100 ° + cos 60 ° = 1 8 cos 80 ° + 1 8 cos 100 ° + 1 8 cos 60 ° = 1 8 cos 80 ° + cos 100 ° + 1 8 1 2 = 1 8 2 cos 80 ° + 100 ° 2 cos 80 ° - 100 ° 2 + 1 16 = 1 8 2 cos 90 ° cos 10 ° + 1 16 = 0 + 1 16 = 1 16 = RHS Hence, proved.

Prove the following: sin 20 ° sin 40 ° sin 60 ° sin 80 ° = 3 16

LHS = sin 20 ° sin 40 ° sin 60 ° sin 80 ° = 3 2 sin 80 ° sin 40 ° sin 20 ° ∵ sin 60 ° = 3 2 = 3 4 2 sin 80 ° sin 40 ° sin 20 ° = 3 4 cos 80 ° - 40 ° - cos 80 ° + 40 ° sin 20 ° ∵ 2 sin A sin B = cos A - B - cos A + B = 3 4 cos 40 ° - cos 120 ° sin 20 ° = 3 4 cos 40 ° - cos 90 + 30 ° sin 20 ° = 3 4 cos 40 ° + sin 30 ° sin 20 ° ∵ cos 90 ° + x = - sin x = 3 4 cos 40 ° + 1 2 sin 20 ° = 3 4 × 2 2 cos 40 ° + 1 sin 20 ° = 3 4 × 2 2 cos 40 ° sin 20 ° + sin 20 ° = 3 8 sin 40 ° + 20 ° - sin 40 ° - 20 ° sin 20 ° ∵ 2 cos A sin B = sin A + B - sin A - B = 3 8 sin 60 ° - sin 20 ° + sin 20 ° = 3 8 × 3 2 = 3 16 = RHS Hence, proved.

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Mathematics and Statistics (Arts & Science) Part 1>Chapter 3 - Trigonometry - II>EXERCISE 3.4

Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 4: EXERCISE 3.4

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 4: EXERCISE 3.4

Attempt the practice questions on Chapter 3: Trigonometry - II, Exercise 4: EXERCISE 3.4 with hints and solutions to strengthen your understanding. Mathematics and Statistics (Arts & Science) Part 1 solutions are prepared by Experienced Embibe Experts.