Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
Question.

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

solution:

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

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

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

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

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

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

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

$=\mathrm{R} . \mathrm{H} \mathrm{S} .$
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