Prove each of these identities: sin A cos B + cos A sin B ≡ 2 cos ( A - B ) sin 2 B

To prove: sin A cos B + cos A sin B ≡ 2 cos ( A - B ) sin 2 B LHS : sin A cos B + cos A sin B ≡ sin A sin B + cos A cos B sin B cos B ≡ 2 cos A - B 2 sin B cos B ≡ 2 cos A - B sin 2 B = RHS Hence, proved.

Prove the identities cos A + sin A cos A - sin A ≡ sec 2 A + tan 2 A

Given that RHS = sec 2 A + tan 2 A We know that, cos 2 A = 1 - tan 2 A 1 + tan 2 A and tan 2 A = 2 tan A 1 - tan 2 A . ⇒ RHS = 1 + tan 2 A 1 - tan 2 A + 2 tan A 1 - tan 2 A = 1 + tan 2 A + 2 tan A 1 - tan 2 A = 1 + tan A 2 1 - tan A 1 + tan A = 1 + tan A 1 - tan A = 1 + sin A cos A 1 - sin A cos A = cos A + sin A cos A - sin A = LHS Hence, proved.

Prove each of these identities. 1 - tan 2 A 1 + tan 2 A ≡ cos 2 A

To prove: 1 - tan 2 A 1 + tan 2 A ≡ cos 2 A LHS : 1 - tan 2 A 1 + tan 2 A ≡ 1 - sin 2 A cos 2 A 1 + sin 2 A cos 2 A ≡ cos 2 A - sin 2 A cos 2 A + sin 2 A ≡ cos 2 A 1 ≡ cos 2 A = RHS Hence, proved.

Prove the identity: cos 2 A + sin 2 A - 1 cos 2 A - sin 2 A + 1 = tan A

Given that, LHS = cos 2 A + sin 2 A - 1 cos 2 A - sin 2 A + 1 We know that, cos 2 A = 2 cos 2 A - 1 and sin 2 A = 2 sin A cos A = 2 cos 2 A + 2 sin A cos A - 2 2 cos 2 A - 2 sin A cos A We know that, sin 2 A + cos 2 A = 1 = - sin 2 A + sin A cos A cos 2 A - sin A cos A = - sin A sin A - cos A - cos A sin A - cos A = tan A = RHS Hence, proved.

Prove the identity: sin 3 A + sin A 2 sin 2 A ≡ cos A

To prove: sin 3 A + sin A 2 sin 2 A ≡ cos A LHS : sin 3 A + sin A 2 sin 2 A ≡ 3 sin A - 4 sin 3 A + sin A 2 × 2 × sin A cos A ≡ 1 - sin 2 A sin A cos A ≡ cos 2 A sin A cos A ≡ cos A = RHS Hence, proved.

Prove the identity: cos 3 A - sin 3 A 1 - 2 sin 2 A = cos A + sin A .

Given that LHS = cos 3 A - sin 3 A 1 - 2 sin 2 A We know that, cos 3 A = 4 cos 3 A - 3 cos A , sin 3 A = 3 sin A - 4 sin 3 A and sin 2 A = 2 sin A cos A . ⇒ LHS = 4 cos 3 A - 3 cos A - 3 sin A - 4 sin 3 A 1 - 2 2 sin A cos A = 4 cos 3 A - 3 cos A - 3 sin A + 4 sin 3 A 1 - 4 sin A cos A = 4 cos A 1 - sin 2 A - 3 cos A - 3 sin A + 4 sin A 1 - cos 2 A 1 - 4 sin A cos A = cos A + sin A - 4 sin A cos A sin A + cos A 1 - 4 sin A cos A = 1 - 4 sin A cos A sin A + cos A 1 - 4 sin A cos A = sin A + cos A = RHS Hence, proved.

Prove the identity: cos 2 A + 9 cos A + 5 4 + cos A = 2 cos A + 1 .

Given that, LHS : cos 2 A + 9 cos A + 5 4 + cos A We know that, cos 2 A = 2 cos 2 A - 1 . = 2 cos 2 A - 1 + 9 cos A + 5 4 + cos A = 2 cos 2 A + 9 cos A + 4 4 + cos A = ( 2 cos A + 1 ) ( cos A + 4 ) 4 + cos A = 2 cos A + 1 = RHS Hence, proved.

Prove the identity: cos 3 A - sin 3 A cos A - sin A = 2 + sin 2 A 2

Given that LHS = cos 3 A - sin 3 A cos A - sin A = ( cos A - sin A ) cos 2 A + sin 2 A + cos A sin A cos A - sin A We know that, sin 2 A + cos 2 A = 1 . ⇒ LHS = 1 + cos A sin A Multiply and divide with 2 we get = 2 1 + cos A sin A 2 We know that, sin 2 A = 2 sin A cos A = 2 + sin 2 A 2 = RHS Hence, proved.

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Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>EXERCISE 3D

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 5: EXERCISE 3D

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 5: EXERCISE 3D

Attempt the free practice questions on Chapter 3: Trigonometry, Exercise 5: EXERCISE 3D with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.