Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
It is known that $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$
$=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\left(\frac{3 \pi}{4}-x\right)\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}$
$=-2 \sin \left(\frac{3 \pi}{4}\right) \sin x$
$=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x$
$=-2 \sin \frac{\pi}{4} \sin x$
$=-2 \times \frac{1}{\sqrt{2}} \times \sin x$
$=-\sqrt{2} \sin x$
$= R . H.S.$
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