Question

Find the value of ${\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{5\pi }{8}+{\cos }^4\frac{7\pi }{8}$.

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

Answer 1

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

${\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{5\pi }{8}+{\cos }^4\frac{7\pi }{8}$

$={\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4(\pi -\frac{3\pi }{8})+{\cos }^4(\pi -\frac{\pi }{8})$

$={\cos }^4\frac{\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{3\pi }{8}+{\cos }^4\frac{\pi }{8}$

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

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

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

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

$=2\left[\right(1{)}^2-2{\sin }^2\frac{3\pi }{8}{\cos }^2\frac{3\pi }{8}]$

$=2-4{\sin }^2\frac{3\pi }{8}{\cos }^2\frac{3\pi }{8}$

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

$=2-{\left(\sin \frac{3\pi }{4}\right)}^2$

$=2-{\left[\sin \right(\pi -\frac{\pi }{4}\left)\right]}^2$

$=2-{\left[\sin \frac{\pi }{4}\right]}^2$

$=2-{\left[\frac{1}{\sqrt{2}}\right]}^2$

$=2-\frac{1}{2}$

$=\frac{3}{2}$ (Option D)