Question

The value of the expression ${\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{5\pi }{8}+{\cos }^4\frac{7\pi }{8}$ is

• A. $2$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

$\because \frac{\pi }{8},\frac{7\pi }{8}$ and $\frac{3\pi }{8},\frac{5\pi }{8}$ are
Supplementary angles
$\therefore \cos \frac{7\pi }{8}=\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}$
Similarly,
$\cos \frac{5\pi }{8}=-\cos \frac{3\pi }{8}$
Therefore,
${\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{5\pi }{8}+{\cos }^4\frac{7\pi }{8}=2[{\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}]$

Also, $\frac{\pi }{8}$ and $\frac{3\pi }{8}$ are complementary angles
$\therefore \cos \frac{3\pi }{8}=\sin \frac{\pi }{8}=2[{\cos }^4\frac{\pi }{8}+{\sin }^4\frac{\pi }{8}]=2[{({\cos }^2\frac{\pi }{8}+{\sin }^2\frac{\pi }{8})}^2-2{\cos }^2\frac{\pi }{8}{\sin }^2\frac{\pi }{8}]=2[1-\frac{1}{2}{\left(2\cos \frac{\pi }{8}\sin \frac{\pi }{8}\right)}^2]=2[1-\frac{1}{2}{\sin }^2\frac{\pi }{4}]=2[1-\frac{1}{4}]=\frac{3}{2}$