Question

If $8x=\pi$, show that $\cos 7x+\cos x=0$.

Answer 1

Given:$8x=\pi$

$\Rightarrow x=\frac{\pi }{8}$

L.H.S$=\cos 7x+\cos x$

$=\cos \frac{7\pi }{8}+\cos \frac{\pi }{8}$

$=\cos (\pi -\frac{\pi }{8})+\cos \frac{\pi }{8}$

$=-\cos \frac{\pi }{8}+\cos \frac{\pi }{8}=0=$R.H.S since $\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}$ and $\pi -\theta$ is in second quadrant where $\cos (\pi -\theta )$ is negative

Hence proved.