# Thread: Problem #20: Sum of Sines

1. ## Problem #20: Sum of Sines

We haven't done a Trigonometry problem in a while, so here we go!

Evaluate: $\displaystyle sin^4 \left ( \frac{\pi}{8} \right ) + sin ^4 \left ( \frac{3 \pi}{8} \right ) + sin ^4 \left ( \frac{5 \pi}{8} \right ) + sin ^4 \left ( \frac{7 \pi}{8} \right )$

-Dan

2. ## Re: Problem #20: Sum of Sines

Do you know any half angle formulas? Do you know what $\displaystyle sin(\pi/2)$? So what are $\displaystyle sin(\pi/4)$ and $\displaystyle sin(\pi/8)$? Once you have that the rest is easy.

3. ## Re: Problem #20: Sum of Sines

That serves as a good hint.

-Dan

4. ## Re: Problem #20: Sum of Sines

No final takers? Here's my solution.

Spoiler:
I prefer a slightly different approach than that given by HallsofIvy:

$\displaystyle cos(2 \theta ) = 1 - 2~sin^2( \theta ) \implies sin^4( \theta) = \left ( \frac{ 1 - 2~cos( 2 \theta )}{2} \right )^2$

So for $\displaystyle \theta = \frac{\pi}{8}$ we get
$\displaystyle sin^4 \left ( \frac{\pi}{8} \right ) = \left ( \frac{ 1 - 2~cos \left ( 2 \cdot \frac{\pi}{8} \right ) }{2} \right )^2 = \left ( \frac{ 1 - 2~cos \left ( \frac{\pi}{4} \right ) }{2} \right )^2 = \frac{3}{8} - \frac{\sqrt{2}}{4}$

Similarly
$\displaystyle sin^4 \left ( \frac{3 \pi}{8} \right ) = \frac{3}{8} + \frac{\sqrt{2}}{4}$

$\displaystyle sin^4 \left ( \frac{5 \pi}{8} \right ) = \frac{3}{8} + \frac{\sqrt{2}}{4}$

$\displaystyle sin^4 \left ( \frac{7 \pi}{8} \right ) = \frac{3}{8} - \frac{\sqrt{2}}{4}$