Question

The value of ${\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{5\pi }{8}\right)+{\cos }^4\left(\frac{7\pi }{8}\right)$ is

• A. $0$
• B. $1/2$
• C. $3/2$
• D. $1$

Answer 1

Answer: (C)

$3/2$

Since $\frac{1+\cos 2\theta }{2}={\cos }^2\theta$

$\Rightarrow {\cos }^4\left(\frac{\pi }{8}\right)={\left[\frac{1+\cos \left(2\pi /8\right)}{2}\right]}^2={\left[\frac{1+\cos \left(\pi /4\right)}{2}\right]}^2$
Similarly
${\cos }^4\frac{3\pi }{8}={\left[\frac{1+\cos \left(3\pi /4\right)}{2}\right]}^2$
${\cos }^4\frac{5\pi }{8}={\left[\frac{1+\cos \left(5\pi /4\right)}{2}\right]}^2$
${\cos }^4\frac{7\pi }{8}={\left[\frac{1+\cos \left(7\pi /4\right)}{2}\right]}^2$
$\therefore$ Given expression will become
$\frac{1}{4}[2{(1+\frac{1}{\sqrt{2}})}^2+2{(1+\frac{1}{\sqrt{2}})}^2]=\frac{1}{4}\times 2(2+1)=\frac{1}{4}\times 2\times 3=\frac{3}{2}$