Question

Prove that $\cos \left[\left(\frac{3\mathrm{\pi }}{4}\right)+x\right]-\cos \left[\left(\frac{3\mathrm{\pi }}{4}\right)-x\right]=-\surd 2\sin x.$

Answer 1

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

We know that, $\cos \left(a+b\right)-\cos \left(a-b\right)=-2\sin a\sin b$

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

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

$=-2.\sin \frac{\mathrm{\pi }}{4}\sin x$………………………..(since, $\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\sin \theta$)

$=-2\times \frac{1}{\sqrt{2}}\times \sin x$……………………($\sin \frac{\mathrm{\pi }}{4}=\sin 45°=\frac{1}{\sqrt{2}}$)

$=-\sqrt{2}\sin x$……………..(As, $\sqrt{2}\times \sqrt{2}=2$)

$=R.H.S$

Therefore, it is proved that $\cos \left[\left(\frac{3\mathrm{\pi }}{4}\right)+x\right]-\cos \left[\left(\frac{3\mathrm{\pi }}{4}\right)-x\right]=-\surd 2\sin x.$