Trigonometry/The summation of finite series

Problem Statement

Find a closed form for

$\sin (a)+\sin (a+b)+\sin (a+2b)+\cdots +\sin (a+(n-1)b)$

Note: A 'closed form' is not mathematically defined, but just means a simplified formula which does not involve '...', or a summation sign. In our problem, we should look for a formula that only involves variables $a,b,n$, and known operations like the four operations, radicals, exponents, logarithm, and trigonometric functions.

Method 1

To sum the series

$\sin (a)+\sin (a+b)+\sin (a+2b)+\cdots +\sin (a+(n-1)b)=S$

Multiply each term by

$2\sin \left(\frac{b}{2}\right)$

Then we have

$2\sin (a)\sin \left(\frac{b}{2}\right)=\cos (a-\frac{b}{2})-\cos (a+\frac{b}{2})$

and similarly for all terms to

$2\sin (a+(n-1)b)\sin \left(\frac{b}{2}\right)=\cos (a+(2n-3)\frac{b}{2})-\cos (a+(2n-1)\frac{b}{2})$

Summing, we find that nearly all the terms cancel out and we are left with

$2S\sin \left(\frac{b}{2}\right)=\cos (a-\frac{b}{2})-\cos (a+(2n-1)\frac{b}{2})=2\sin (a+(n-1)\frac{b}{2})\sin \left(\frac{nb}{2}\right)$

Hence

$S=\sin (a+(n-1)\frac{b}{2})\frac{\sin \left(n\frac{b}{2}\right)}{\sin \left(\frac{b}{2}\right)}$

Similarly, if

$C=\cos (a)+\cos (a+b)+\cos (a+2b)+\cdots +\cos (a+(n-1)b)$

then

$C=\cos (a+(n-1)\frac{b}{2})\frac{\sin \left(n\frac{b}{2}\right)}{\sin \left(\frac{b}{2}\right)}$

Method 2

Consider the following sum

$s=e^{ai}+e^{(a+b)i}+\cdots +e^{(a+(n-1)b)i}$

Since $s$ is a geometric series with common ratio $e^{bi}$, we get

$s=\frac{e^{ai}(e^{nbi}-1)}{e^{bi}-1}=\frac{e^{ai}(e^{nbi}-1)}{e^{bi}-1}=\frac{e^{ai}{e}^{\frac{nbi}{2}}(e^{\frac{nbi}{2}}-e^{-\frac{nbi}{2}})}{e^{\frac{bi}{2}}(e^{\frac{bi}{2}}-e^{-\frac{bi}{2}})}$

$s=\frac{\sin \left(n\frac{b}{2}\right)}{\sin \left(\frac{b}{2}\right)}\cdot e^{(a+(n-1)\frac{b}{2})i}$

Therefore,

$\sin (a)+\sin (a+b)+\sin (a+2b)+\cdots +\sin (a+(n-1)b)=\mathrm{I}\mathrm{m}(s)=\sin (a+(n-1)\frac{b}{2})\frac{\sin \left(n\frac{b}{2}\right)}{\sin \left(\frac{b}{2}\right)}$

$\cos (a)+\cos (a+b)+\cos (a+2b)+\cdots +\cos (a+(n-1)b)=\mathrm{R}\mathrm{e}(s)=\cos (a+(n-1)\frac{b}{2})\frac{\sin \left(n\frac{b}{2}\right)}{\sin \left(\frac{b}{2}\right)}$