$\sin \frac{2 \pi}{5}+\sin \frac{4 \pi}{5}+\sin \frac{6 \pi}{5}+\sin \frac{8 \pi}{5}$ is equal to
$\sin \frac{2 \pi}{5}+\sin \frac{4 \pi}{5}+\sin \frac{6 \pi}{5}+\sin \frac{8 \pi}{5}$ is equal to
- 0
- 1
- $\frac{\sqrt{2}}{2}$
- $\frac{1}{2}$
Solution
$\sin \frac{2 \pi}{5}+\sin \frac{4 \pi}{5}+\sin \frac{6 \pi}{5}+\sin \frac{8 \pi}{5}$
$=\left(\sin \frac{2 \pi}{5}+\sin \frac{8 \pi}{5}\right)+\left(\sin \frac{4 \pi}{5}+\sin \frac{6 \pi}{5}\right)$
$
\begin{aligned}
& =2 \sin \left(\frac{10 \pi}{5 \times 2}\right) \cos \left(\frac{-6 \pi}{5 \times 2}\right)+2 \sin \left(\frac{10 \pi}{5 \times 2}\right) \cos \left(\frac{-2 \pi}{5 \times 2}\right) \\
& =2 \sin \pi \cos \left(-\frac{3 \pi}{5}\right)+2 \sin \pi \cos \left(\frac{-2 \pi}{10}\right)[\because \sin \pi=0] \\
& =0
\end{aligned}
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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