સાબિત કરો કે : $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$

It is known that $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)$

$\therefore$ $L.H.S.$ $=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$

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

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

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

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

$=-2 \times \frac{1}{\sqrt{2}} \times \sin x$

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

$= R . H.S.$