Simplify a quick sum of sines

Question

Simplify $\sin 2+\sin 4+\sin 6+\cdots+\sin 88$

I tried using the sum-to-product formulae, but it was messy, and I didn't know what else to do. Could I get a bit of help? Thanks.

Answer 1

If you know about complex numbers, you may want to use that: \begin{array}{lcl}\sum_{n=1}^N\sin 2n & = &\Re\left(\sum_{n=1}^N\left(\sin 2n+i\cos 2n\right)\right) \\ & = & \Re\left(\sum_{n=1}^Ne^{2ni}\right) \\ & = & \Re\left(e^{2i}\frac{1-e^{2Ni}}{1-e^{2i}}\right).\end{array}

Answer 2

The angles are in arithmetic progression. Use the formula

$$\sum_{k=0}^{n-1} \sin (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \sin \left( a+ (n-1)\frac{b}{2}\right)$$

See here for two proofs (using trigonometry, or using complex numbers).

In your case, $a=b=2$ and $n=44$.

Answer 3

$$\sum_{n=1}^{N} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(N+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$

$$\sum_{n=1}^{44} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(44+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$

$$\sum_{n=1}^{44} {\sin(2 n)} = \frac{1}{2} \cot{\frac{2}{2}} - \frac{\cos{[(44+\frac{1}{2})2]}}{2\sin{\frac{2}{2}}}$$

$$\sum_{n=1}^{44} {\sin(2 n)} = \frac{1}{2} \cot{1} - \frac{1}{2}\frac{\cos{99}}{\sin{1}}$$

$$\sum_{n=1}^{44} {\sin(2 n)} \approx 0.2974$$

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities