Question

Prove the following identities
$2f^{\prime }(x+\frac{\pi }{3})f^{\prime }(x-\frac{\pi }{6})=f^{\prime }\left(0\right)-f(2x+\frac{\pi }{6})$, where $f\left(x\right)=\cos x$

Answer 1

$2f^{\prime }(x+\frac{\pi }{3})f^{\prime }(x-\frac{\pi }{6})=f^{\prime }\left(0\right)-f(2x+\frac{\pi }{6})$ where $f\left(x\right)=\cos x$

$f\left(x\right)=\cos x\Rightarrow f^{\prime }\left(x\right)=-\sin x$

$f^{\prime }(x+\frac{\pi }{3})=-\sin (x+\frac{\pi }{3})=-(\sin x\cos \frac{\pi }{3}+\cos x\sin \frac{\pi }{3})$

$=-(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)$

$=\frac{-1}{2}[\sin +\sqrt{3}\cos x]$

$f^{\prime }(x-\frac{\pi }{6})=-\sin (x-\frac{\pi }{6})=-(\sin x\cos \frac{\pi }{6}-\cos x\sin \frac{\pi }{6})$

$=-(\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x)=\frac{-1}{2}(\sqrt{3}\sin x-\cos x)$

$2f^{\prime }(x+\frac{\pi }{3})f^{\prime }(x-\frac{\pi }{6})=\frac{1}{2}\left[\right(\sin x+\sqrt{3}\cos x\left)\right(\sqrt{3}\sin x-\cos x\left)\right]$

$=\frac{1}{2}[\sqrt{3}{\sin }^{2}x+2\sin x\cos x-\sqrt{3}{\cos }^{2}x]$

$=\frac{1}{2}[\sin 2x-\sqrt{3}\cos 2x]$

Now,

$f^{\prime }\left(0\right)=0$

$f(2x+\frac{\pi }{6})=\cos (2x+\frac{\pi }{6})=(\cos 2x\cos \frac{\pi }{6}-\sin 2x\sin \frac{\pi }{6})=(\frac{\sqrt{3}}{2}\cos 2x-\frac{1}{2}\sin 2x)$

$f^{\prime }\left(0\right)-f(2x+\frac{\pi }{6})=\frac{-1}{2}(\sqrt{3}\cos 2x-\sin 2x)=\frac{1}{2}(\sin 2x-\sqrt{3}\cos 2x)$

$LHS=RHS$