Trigonometry/The summation of finite series

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Problem StatementEdit

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 1Edit

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)a\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

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

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

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

then

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 2Edit

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