Find the sum $\sum_{n=0}^{49} \sin((2n+1)x) $

Question

This was an exercise in a chapter of a textbook on product to sum and sum to product trigonometric identities. The following question was asked with the given hint: $$\sum_{n=0}^{49} \sin((2n+1)x) $$ Hint: multiply this sum by $2\sin(x)$

My attempt

$$\sum_{n=0}^{49} \sin((2n+1)x)=1/2\csc(x)\sum_{n=0}^{49} 2\sin(x)\sin((2n+1)x) $$ Using identity $2\sin(A)\sin(B)=\cos(A-B)-\cos(A+B)$ $$1/2\csc(x)\sum_{n=0}^{49} 2\sin(x)\sin((2n+1)x)=1/2\csc(x)\sum_{n=0}^{49} \cos(2nx)-\cos((2n+2)x)$$ How do I continue from here?

Your final sum is telescoping. Write out the first few terms you should see that it simplifies to $$1-\cos{(100x)}$$ — Peter Foreman, Jul 19 '20 at 07:55
https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Jul 19 '20 at 08:00

score 2 · Accepted Answer · answered Jul 19 '20 at 08:04

$$S_n=\sum_{0}^{n} \sin ((2k+1)x)=\Im \sum_{k=0}^{n} e^{i(2k+1)x}=\Im e^{ix}\sum_{k=0}^{n} e^{2ikx} =\Im e^{ix} \frac{e^{2inx}-1}{e^{2ix}-1}=\Im e^{i)n+1)x} \frac{e^{i)n+1)x}-e^{-i(n+1)x}}{e^{ix}-e^{-ix}}$$ $$\implies S_n=\frac{\sin^2 (n+1)x}{\sin x}$$ So finally, $$S_{49}=\frac{\sin^2 50 x}{\sin x}$$

Sameer Baheti · Answer 2 · 2020-07-19T09:16:25.670

0

These are very common summations in sequences and series in which only the terms on either ends remain. \begin{align*} &\Rightarrow\sum_{n=0}^{m} f(2nx)-f((2n+2)x)\\ &=\color{red}{f(0)}-f(2x)+f(2x)-f(4x)+\cdots -f(2mx)+f(2mx)-\color{blue}{f((2m+2)x)}\\ &=\color{red}{f(0)}-\color{blue}{f((2m+2)x)} \end{align*} where $f=\cos$ and $m\ge n$.

So, $$S=\frac{1-\cos(100x)}{2\sin x}=\frac{\sin^2 50 x}{\sin x}$$

edited Jul 19 '20 at 09:16

answered Jul 19 '20 at 08:07

Sameer Baheti

3,239

Find the sum $\sum_{n=0}^{49} \sin((2n+1)x) $

My attempt

2 Answers2