Prove $\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$ Please help me to solve the below trigonometric function as i am trying it from the last hour.
$$\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$$
 A: Apply, $\cos(A+B)-\cos(A-B)=-2\sin A \sin B$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-2\sin x\sin(3\frac{\pi}{4}) $
$=-2\sin x\sin(\pi-\frac{\pi}{4})=-2\sin x\sin(\frac{\pi}{4})$
 as $\sin(\pi-y)=\sin y$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-\sqrt2\sin x$
or apply $\cos 2C- \cos 2D=-2sin(C+D)sin(C-D)$
Put $2C=3\frac{\pi}{4}+x, 2D=3\frac{\pi}{4}-x=>C+D=3\frac{\pi}{4}$ and $C-D=x$
A: Probably you are meant to use the formulas $\cos(a+b)=\cos a\cos b-\sin a\sin b$ and $\cos(a-b)=\cos a\cos b+\sin a\sin b$. We could let $a=3\pi/4$ and $b=x$, and compute, then subtract, using the fact that $\cos(3\pi/4)=-1/\sqrt{2}$ and $\sin(3\pi/4)=1/\sqrt{2}$.  
But it is much better to subtract first, as lab bhattacharjee did, because there is very nice cancellation, and we get 
$$(\cos a\cos b-\sin a\sin b)-(\cos a\cos b+\sin a\sin b)=-2\sin a\sin b.$$
Now put $a=3\pi/4$ and $b=x$. Since $\sin(3\pi/4)=1/\sqrt{2}$, we get
$$-\frac{2}{\sqrt{2}}\sin x.$$
Finally, note that $\frac{2}{\sqrt{2}}=\sqrt{2}$.
