Q67 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

If $f(\theta)=\cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\theta-\frac{2 \pi}{3}\right)$ then $f\left(\frac{\pi}{5}\right)=$

  • A
    $\frac{3(\sqrt{5}-1)}{16}$
  • B
    $\frac{3 \sqrt{10-2 \sqrt{5}}}{8}$
  • C
    $\frac{3 \sqrt{10+2 \sqrt{5}}}{8}$
  • Correct answer
    D
    $\frac{3(\sqrt{5}+1)}{16}$

Solution

$\begin{aligned} & \quad f(\theta)=\cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\theta-\frac{2 \pi}{3}\right) \\ & \cos \left(\theta-\frac{2 \pi}{3}\right)=\cos \left(\pi-\left(\frac{\pi}{3}+\theta\right)\right)=\cos \left(\pi+\frac{\pi}{3}+\theta\right) \\ & =\cos \left(\frac{4 \pi}{3}+\theta\right) \\ & \because \cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\frac{4 \pi}{3}+\theta\right)=\frac{3}{4} \cos 3 \theta \\ & f(\theta)=\frac{3}{4} \cos 3 \theta \\ & f\left(\frac{\pi}{5}\right)=\frac{3}{4} \cos \frac{3 \pi}{5}=\frac{3}{4}\left(\frac{\sqrt{5}+1}{4}\right)=\frac{3}{16}(\sqrt{5}+1) .\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions