Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C

$\frac{3}{2}$

Explanation for the correct answer:

Simplifying the equations using trigonometric identities:

GIven expression is: ${\sin }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{5\pi }{8}\right)+{\sin }^4\left(\frac{7\pi }{8}\right)$

Simplifying the above expression:

$$\begin{gathered} ={\sin }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{5\pi }{8}\right)+{\sin }^4\left(\frac{7\pi }{8}\right) \\ ={\left[{\sin }^2\left(\frac{\pi }{8}\right)\right]}^2+{\left[{\sin }^2\left(\frac{3\pi }{8}\right)\right]}^2+{\left[{\sin }^2\left(\frac{5\pi }{8}\right)\right]}^2+{\left[{\sin }^2\left(\frac{7\pi }{8}\right)\right]}^2 \\ ={\left(\frac{1–\cos {\displaystyle \frac{\pi }{4}}}{2}\right)}^2+{\left(\frac{1–\cos {\displaystyle \frac{3\pi }{4}}}{2}\right)}^2+{\left(\frac{1–\cos \frac{3\pi }{4}}{2}\right)}^2+{\left(\frac{1–\cos {\displaystyle \frac{7\pi }{4}}}{2}\right)}^2\left[\because {\sin }^{2}A=\frac{1–\cos 2A}{2}\right] \\ ={\left(\frac{1-\frac{\sqrt{2}}{2}}{2}\right)}^2+{\left(\frac{1+\frac{\sqrt{2}}{2}}{2}\right)}^2+{\left(\frac{1-\frac{\sqrt{2}}{2}}{2}\right)}^2+{\left(\frac{1+\frac{\sqrt{2}}{2}}{2}\right)}^2\left[\because \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\right] \\ =2\left[{\left(\frac{1-\frac{\sqrt{2}}{2}}{2}\right)}^2\right]+2\left[{\left(\frac{1+\frac{\sqrt{2}}{2}}{2}\right)}^2\right] \\ =2{\left[\frac{2-\sqrt{2}}{4}\right]}^2+2{\left[\frac{2+\sqrt{2}}{4}\right]}^2 \\ =\frac{1}{8}{\left(2-\sqrt{2}\right)}^2+\frac{1}{8}{\left(2+\sqrt{2}\right)}^2 \\ =\frac{1}{8}\left[\left(4-4\sqrt{2}+2\right)+\left(4+4\sqrt{2}+2\right)\right] \\ =\frac{12}{8} \\ =\frac{3}{2} \end{gathered}$$

Therefore, the correct answer is Option (C).