Question 6 & 7, Review Exercise 10

Solutions of Question 6 & 7 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove the identity $\cos 4\theta =1-8{{\sin }^{2}}\theta {{\cos }^{2}}\theta $.

\begin{align}L.H.S&=\cos 4\theta \\ &=\cos 2\left( 2\theta \right)\\ &=1-2si{{n}^{2}}2\theta \\ &=1-2{{\left( 2sin\theta \cos \theta \right)}^{2}}\\ &=1-8si{{n}^{2}}\theta {{\cos }^{2}}\theta =R.H.S.\end{align}

Prove the identity $\sin 6x\sin x+\cos 4x\cos 3x=\cos 3x\cos 2x$.

\begin{align}L.H.S.&=\sin 6x\sin x+\cos 4x\cos 3x\\ &=\dfrac{1}{2}\left( 2\sin 6x\sin x+2\cos 4x\cos 3x \right)\\ &=\dfrac{1}{2}\left[ \cos \left( 6x-x \right)-\cos \left( 6x+x \right)+\left( \cos \left( 4x+3x \right)+\cos \left( 4x-3x \right) \right) \right]\\ &=\dfrac{1}{2}\left[ \cos 5x-\cos 7x+\left( \cos 7x+\cos x \right) \right]\\ &=\dfrac{1}{2}\left[ \cos 5x-\cos 7x+\cos 7x+\cos x \right]\\ &=\dfrac{1}{2}\left[ \cos 5x+\cos x \right]\\ &=\dfrac{1}{2}\left[ 2\cos \dfrac{5x+x}{2}+\cos \dfrac{5x-x}{2} \right]\\ &=\dfrac{1}{2}\left( 2\cos 3x+\cos 2x \right)\\ \cos 3x+\cos 2x&=R.H.S.\end{align}

< Question 4 & 5 Question 8 & 9 >