Question

Evaluate:
${\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{5\pi }{8}\right)+{\cos }^4\left(\frac{7\pi }{8}\right)$

Answer 1

Let $A={\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{5\pi }{8}\right)+{\cos }^4\left(\frac{7\pi }{8}\right)$

${\cos }^4\left(\frac{5\pi }{8}\right)={\cos }^4\left(\frac{\pi }{2}+\frac{\pi }{8}\right)={\sin }^4\left(\frac{\pi }{8}\right)$

${\cos }^4\left(\frac{7\pi }{8}\right)={\cos }^4\left(\frac{\pi }{2}+\frac{3\pi }{8}\right)={\sin }^4\left(\frac{3\pi }{8}\right)$

$\therefore A={\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)$

Let's write in the form, $(a+b{)}^2=a^2+2ab+b^2$

$={\left({\sin }^2\left(\frac{\pi }{8}\right)+{\cos }^2\left(\frac{\pi }{8}\right)\right)}^2+{\left({\sin }^2\left(\frac{3\pi }{8}\right)+{\cos }^2\left(\frac{3\pi }{8}\right)\right)}^2-2{\sin }^2\left(\frac{\pi }{8}\right){\cos }^2\left(\frac{\pi }{8}\right)-2{\sin }^2\left(\frac{3\pi }{8}\right){\cos }^2\left(\frac{3\pi }{8}\right)$

$=1+1-\frac{1}{2}{\sin }^2\frac{\pi }{4}-\frac{1}{2}{\sin }^2\frac{3\pi }{4}$ .................. $[\because \sin 2x=2\sin x\cos x]$ and $[sin²\theta +cos²\theta =1]$

$=2-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}\right)=2-\frac{1}{2}=\frac{3}{2}$