====== 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.
=====Question 6=====
Prove the identity $\cos 4\theta =1-8{{\sin }^{2}}\theta {{\cos }^{2}}\theta $.
====Solution====
\begin{align}L.H.S&=\cos 4\theta \\
&=\cos 2\left( 2\theta  \right)\\
&=1-2\sin^2 2\theta \\
&=1-2{{\left( 2\sin\theta \cos \theta  \right)}^{2}}\\
&=1-8\sin^2 \theta {{\cos }^{2}}\theta =R.H.S.\end{align}

=====Question 7=====
Prove the identity $\sin 6x\sin x+\cos 4x\cos 3x=\cos 3x\cos 2x$.
====Solution====
\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} 
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