Problem Statement

Find a closed form for

${\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)}$ .

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 ${\displaystyle a,b,n}$ , and known operations like the four operations, radicals, exponents, logarithm, and trigonometric functions.

Method 1

To sum the series

${\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)=S}$ .

Multiply each term by

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

Then we have

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

and similarly for all terms to

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

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

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

Hence

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

Similarly, if

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

then

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

Method 2

Consider the following sum

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

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

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

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

Therefore,

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

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