Question

${\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{5\pi }{8}+{\cos }^4\frac{7\pi }{8}=\frac{m}{2}$. Find $m$

Answer 1

We have $\frac{7\pi }{8}=\pi -\frac{\pi }{8}$ and $\frac{5\pi }{8}=\pi -\frac{3\pi }{8}$
$\Rightarrow$ $\cos \frac{7\pi }{8}=-\cos \frac{\pi }{8}$ and $\cos \frac{5\pi }{8}=-\cos \frac{3\pi }{8}$
$\Rightarrow$ ${\cos }^4\frac{7\pi }{8}=-{\cos }^4\frac{\pi }{8}$ and ${\cos }^4\frac{5\pi }{8}=-{\cos }^4\frac{3\pi }{8}$
$\therefore$ L.H.S. $=2{\cos }^4\frac{\pi }{8}+2{\cos }^4\frac{3\pi }{8}$
$=2[{\left({\cos }^2\frac{\pi }{8}\right)}^2+{\left({\cos }^2\frac{3\pi }{8}\right)}^2]$
$=2{\left\{\frac{1+\cos \frac{\pi }{4}}{2}\right\}}^2+{\left\{\frac{1+\cos \frac{3\pi }{4}}{2}\right\}}^2$ $[\because \frac{1+\cos 2\theta }{2}={\cos }^2\theta ]$
$=\frac{1}{2}{\{{(1+\cos \frac{\pi }{4})}^2+{(1+\cos \frac{\pi }{4})}^2\}}^2$
$=\frac{1}{2}\{{(1+\frac{1}{\sqrt{2}})}^2+{(1-\frac{1}{\sqrt{2}})}^2\}$
$=\frac{1}{2}\{(1+\frac{1}{2}+\sqrt{2})+(1+\frac{1}{2}-\sqrt{2})\}=\frac{3}{2}=$ R.H.S.
Ans: 3