Prove that:

Solution:

(i) Consider LHS :

$\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$

$=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}$

$\left\{\because \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right\}$

$=-2 \sin \frac{3 \pi}{4} \sin x$

$=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x$

$=-2 \sin \frac{\pi}{4} \sin x$

$=-\sqrt{2} \sin x$

(ii) Consider LHS :

$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$

$=2 \cos \left\{\frac{\left(\frac{\pi}{4}+x\right)+\left(\frac{\pi}{4}-x\right)}{2}\right\} \cos \left\{\frac{\left(\frac{\pi}{4}+x\right)-\left(\frac{\pi}{4}+x\right)}{2}\right\}$

$\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos \left\{\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right\} \cos \left\{\frac{\frac{\pi}{4}+x-\frac{\pi}{4}+x}{2}\right\}$

$=2 \cos \left(\frac{\pi}{4}\right) \cos x$

$=2 \times \frac{1}{\sqrt{2}} \times \cos x$

$=\sqrt{2} \cos x$

$=$ RHS

Hence, LHS = RHS