Q60 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics
If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$
- Correct answerA$\frac{2 \pi}{3}$
- B$\frac{3 \pi}{4}$
- C$\frac{\pi}{4}$
- D$\frac{\pi}{2}$
Solution
$\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}$
$=\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4\left(\frac{\pi}{2}+\frac{\pi}{8}\right)+\cos ^4\left(\frac{\pi}{2}+\frac{3 \pi}{8}\right)$
$=\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}$
$=\left(\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{\pi}{8}\right)+\left(\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{3 \pi}{8}\right)$
$\begin{aligned} &=\left(\cos ^2 \frac{\pi}{8}+\sin ^2 \frac{\pi}{8}\right)^2-2 \sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8} \\ &+\left(\cos ^2 \frac{3 \pi}{8}+\sin ^2 \frac{3 \pi}{8}\right)^2-2 \sin ^2 \frac{3 \pi}{8} \cos ^2 \frac{3 \pi}{8}\end{aligned}$
$=2-\frac{1}{2} \sin ^2 \frac{\pi}{4}-\frac{1}{2} \sin ^2 \frac{3 \pi}{4}$
$=2-\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2-\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{3}{2}=k$
Now, $\sin ^{-1} \sqrt{\frac{k}{2}}+\cos ^{-1}\left(\frac{k}{3}\right)=\sin ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1} \frac{1}{2}$
$=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$
Hence, the correct answer is option: 'A'.