Q63 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
$\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{4 \pi}{8}+\cos ^4 \frac{5 \pi}{8}$
$+\cos ^4 \frac{6 \pi}{8}+\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{8 \pi}{8}$ is equal to
- Correct answerA3
- B-1
- C1
- D4
Solution
$\begin{aligned} & \cos ^4 \frac{\pi}{8}+\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{4 \pi}{8} \\ & +\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{6 \pi}{8}+\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{8 \pi}{8} \\ & =\left(\cos ^4 \frac{\pi}{8}+\cos \frac{7 \pi}{8}\right)+\left(\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{6 \pi}{8}\right) \\ & \quad+\left(\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}\right)+\left(\cos ^4 \frac{4 \pi}{8}+\cos ^4 \frac{8 \pi}{8}\right) \\ & =\left(2 \cos ^4 \frac{\pi}{8}+2 \cos ^4 \frac{2 \pi}{8}+2 \cos ^4 \frac{3 \pi}{8}\right)+\cos ^4 \frac{\pi}{2}+\cos ^4(\pi) \\ & =2\left[\cos ^4 \frac{\pi}{8}+\cos ^4\left(\frac{\pi}{2}-\frac{\pi}{8}\right)+\cos ^4 \frac{\pi}{4}\right]+0+1 \\ & =2\left(\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{\pi}{8}+\frac{1}{4}\right)+1 \\ & =2\left\{\left(\cos ^2 \frac{\pi}{8}+\sin ^2 \frac{\pi}{8}\right)^2-2 \sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8}+\frac{1}{4}\right\}+1 \\ & =2\left\{1-2 \sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8}+\frac{1}{4}\right\}+1 \\ & =2-\left(2 \sin ^{\frac{\pi}{8}} \cos ^{\frac{\pi}{8}}\right)^2+\frac{1}{2}+1=3-\left(\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2} \\ & =3\end{aligned}$
Hence, the correct answer is option: 'A'.