Sum of squares of sines.

Question

Any ideas on how to compute this sum? I'm sure there's a simple trick to it, but I just can't wrap my mind around it at the moment. Some insight would be tremendously appreciated, thanks!

$$\sum_{n = 1}^{90} \sin^2(n^\circ) \approx 45.5$$

Answer 1

Compare $\sin(1^{\circ})^2$ and $\sin(89^{\circ})^2$. What connection can you find? Do the same with $\sin(2^{\circ})^2$ and $\sin(88^{\circ})^2$.

Answer 2

Use the fact that $$\sin(z) = {e^z - e^{-iz}\over 2i}.$$ and turn this into a messy geometric sum.

Answer 3

Hint: 1) $\sin^2x = \dfrac{1-\cos 2x}{2}$

2) Find a closed form for the quite popular sum: $\cos 2x + \cos 4x + ....+ \cos 2nx$

Answer 4

Hints: Can you use the following facts to sum your series in pairs?

• $\sin^2 \theta = 1 - \cos^2 \theta$
• $\cos(90^\mathrm o-\theta)=\sin \theta$

For example, $$\sin^2(5^\mathrm o)=1-\cos^2(5^\mathrm o)=1-\sin^2(85^\mathrm o)$$