Prove that:

Question:

Prove that:

$\cos 4 x-\cos 4 \alpha=8(\cos x-\cos \alpha)(\cos x+\cos \alpha)(\cos x-\sin \alpha)(\cos x+\sin \alpha)$

Solution:

RHS $=8(\cos x-\cos \alpha)(\cos x+\cos \alpha)(\cos x-\sin \alpha)(\cos x+\sin \alpha)$

$=8\left(\cos ^{2} x-\cos ^{2} \alpha\right)\left(\cos ^{2} x-\sin ^{2} \alpha\right)$

$=8\left(\cos ^{4} x-\cos ^{2} x \times \sin ^{2} \alpha-\cos ^{2} \alpha \times \cos ^{2} x+\cos ^{2} \alpha \times \sin ^{2} \alpha\right)$

$=8\left\{\cos ^{4} x-\cos ^{2} x\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\cos ^{2} \alpha \times \sin ^{2} \alpha\right\}$

$=8\left\{\cos ^{4} x-\cos ^{2} x+\cos ^{2} \alpha \times\left(1-\cos ^{2} \alpha\right)\right\}$

$=8\left\{\cos ^{4} x-\cos ^{2} x+\cos ^{2} \alpha-\cos ^{4} \alpha\right\}$

$=8\left\{\cos ^{2} x\left(\cos ^{2} x-1\right)+\cos ^{2} \alpha \times\left(1-\cos ^{2} \alpha\right)\right\}$

$=8\left\{\frac{1}{2} \cos ^{2} x\left(2 \cos ^{2} x-2\right)+\frac{1}{2} \cos ^{2} \alpha \times\left(2-2 \cos ^{2} \alpha\right)\right\}$

$=8\left\{\frac{1}{2} \cos ^{2} x\left(2 \cos ^{2} x-1-1\right)-\frac{1}{2} \cos ^{2} \alpha \times\left(2 \cos ^{2} \alpha-1-1\right)\right\}$

$=8\left\{\frac{1}{2} \cos ^{2} x(\cos 2 x-1)-\frac{1}{2} \cos ^{2} \alpha \times(\cos 2 \alpha-1)\right\} \quad\left(\because \cos 2 \alpha=2 \cos ^{2} \alpha-1\right)$

$=8\left[\frac{1}{4}\left\{2 \cos ^{2} x(\cos 2 x-1)-2 \cos ^{2} \alpha \times(\cos 2 \alpha-1)\right\}\right]$

$=8\left[\frac{1}{4}\{(1+\cos 2 x)(\cos 2 x-1)-(1+\cos 2 \alpha)(\cos 2 \alpha-1)\}\right]$

$=8\left[\frac{1}{4}\left\{\cos ^{2} 2 x-1-\cos ^{2} 2 \alpha+1\right\}\right]$

$=8\left[\frac{1}{8}\left\{2 \cos ^{2} 2 x-2 \cos ^{2} 2 \alpha\right\}\right]$

$=[\{(1+\cos 4 x)-(1+\cos 4 \alpha)\}]$

$=[1+\cos 4 x-1-\cos 4 \alpha]$

$=\cos 4 x-\cos 4 \alpha=\mathrm{LHS}$

Hence proved.