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

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

Then we have

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

and similarly for all terms to

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

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

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

Hence

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

Similarly, if

$C=\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots +\cos {\bigl (}a+(n-1)b{\bigr )}$

then

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

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

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

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

Therefore,

$\sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin {\bigl (}a+(n-1)b{\bigr )}={\rm {Im}}(s)=\sin \left(a+(n-1){\tfrac {b}{2}}\right){\frac {\sin \left(n{\tfrac {b}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}$

$\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots +\cos {\bigl (}a+(n-1)b{\bigr )}={\rm {Re}}(s)=\cos \left(a+(n-1){\tfrac {b}{2}}\right){\frac {\sin \left(n{\tfrac {b}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}$