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

Answer 1

The correct option is C

$\frac{3}{2}$

Explanation for correct option

Given trigonometric function is ${\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{5\pi }{8}\right)+{\cos }^4\left(\frac{7\pi }{8}\right)$

Therefore,

$$\begin{gathered} {\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{5\pi }{8}\right)+{\cos }^4\left(\frac{7\pi }{8}\right) \\ ={\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\pi -\frac{3\pi }{8}\right)+{\cos }^4\left(\pi -\frac{\pi }{8}\right) \\ ={\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)+{\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right) \\ =2\left({\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{3\pi }{8}\right)\right) \\ =2\left({\cos }^4\left(\frac{\pi }{8}\right)+{\cos }^4\left(\frac{\pi }{2}-\frac{\pi }{8}\right)\right) \\ =2\left({\cos }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{\pi }{8}\right)\right) \\ =2\left[{\left({\cos }^2\left(\frac{\pi }{8}\right)+{\sin }^2\left(\frac{\pi }{8}\right)\right)}^2-2{\cos }^2\left(\frac{\pi }{8}\right){\sin }^2\left(\frac{\pi }{8}\right)\right]\left[\because a^2+b^2={\left(a+b\right)}^2-2ab\right] \\ =2\left[{\left(1\right)}^2-\frac{1}{2}2.2{\cos }^2\left(\frac{\pi }{8}\right){\sin }^2\left(\frac{\pi }{8}\right)\right] \\ =2\left[1-\frac{1}{2}{\left(2\cos \left(\frac{\pi }{8}\right)\sin \left(\frac{\pi }{8}\right)\right)}^2\right] \\ =2\left[1-\frac{1}{2}{\left(\sin \left(\frac{\pi }{4}\right)\right)}^2\right]\left[\because \sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)\right] \\ =2\left[1-\frac{1}{2}·\frac{1}{2}\right]\left[\because \sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right] \\ =\frac{3}{2} \end{gathered}$$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$