Question

If $f(\theta) = \sin^2 \theta + \sin^2 (\theta + \frac{2\pi}{3}) + \sin^2 (\theta + \frac{4\pi}{3})$, then $f(\frac{\pi}{15})$ is equal to

Answer 1

$\frac{3}{2}$

Answer 2

Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$, we get $f(\theta) = \frac{1 - \cos(2\theta)}{2} + \frac{1 - \cos(2\theta + \frac{4\pi}{3})}{2} + \frac{1 - \cos(2\theta + \frac{8\pi}{3})}{2}$ $f(\theta) = \frac{3}{2} - \frac{1}{2} [\cos(2\theta) + \cos(2\theta + \frac{4\pi}{3}) + \cos(2\theta + \frac{8\pi}{3})]$ Since $\cos(2\theta + \frac{8\pi}{3}) = \cos(2\theta + \frac{2\pi}{3})$, the sum of cosines is $\cos(2\theta) + \cos(2\theta + \frac{2\pi}{3}) + \cos(2\theta + \frac{4\pi}{3}) = 0$. Therefore, $f(\theta) = \frac{3}{2}$.