Prove by induction that $ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$

Question

Prove by induction that

$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ with $n \geq 1$

Testing n=1:

$$\sin(x)= \frac {1-\cos(2x)}{2\sin(x)}$$

$$2\sin^2(x)=1-\cos(2x)$$

$$2\sin^2(x)=1-[\cos^2(x)-\sin^2(x)]$$

$$\sin^2(x)=1-\cos^2(x)$$

$$\sin^2(x)+\cos^2(x) =1$$

It shows that n=1 yields a true identity (Pythagorean identity)

Let's assume that $P_n$ is true: $$\ sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$
Let's consider adding $\sin(2n+1)x$ to $P_n$:

$$ \sin(x) +\sin(3x) +...+ \sin (2n-1)x+ \sin(2n+1)x= \frac{1-\cos(2nx)}{2\sin x} + \sin(2n+1)x$$

Considering the RHS: $$\frac{1-\cos(2nx)}{2\sin x} + \sin (2n+1)x$$ $$\frac{1-\cos(2nx)+ \sin(2n+1)x \cdot 2\sin x}{2 \sin x}$$ $$\frac{1-\cos(2nx)+ 2[\sin(2n+1)x \cdot sin x]}{2sinx}$$ $$\frac{1-\cos(2nx)+ 2 \cdot \frac{1}{2} [\cos[(2n+1)x-x]-\cos[(2n+1)x+x]]}{2\sin x}$$ $$\frac{1-\cos(2nx)+ \cos(2nx)-\cos(2n+2)x}{2\sin x}$$ $$\frac{1-\cos(2n+2)x}{2\sin x}$$

It follows that $$\sin(x) +\sin(3x) +...+ \sin [(2(n+1)-1)x]= \frac{1-\cos(2(n+1)x)}{2\sin x}$$

Therefore, $$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ is true

Any input is much appreciated.

https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, May 27 '17 at 12:34
Looks good. NB: Use \sin and \cos instead of sin and cos in LaTeX. — Zain Patel, May 27 '17 at 12:36
Step $3$ is incorrect. In an induction, you do not assume that $P_{n+1}$ is true, you use the truth of $P_n$ to prove that $P_{n+1}$ is true. Just skip that step, you don't need it. — Michael Burr, May 27 '17 at 12:36
It is good that you show us how you've gotten to your solution, however you've never explicitly stated your question. Would you like us to verify your proof? — projectilemotion, May 27 '17 at 12:39
Possible duplicate of How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? — Simply Beautiful Art, May 27 '17 at 13:05

score 2 · Answer 1 · answered May 27 '17 at 12:41

2

You can also use $$\sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i}$$and geometric series for a direct proof.

answered May 27 '17 at 12:41

user450041

199

I have better. Just use $\sin(x)=\Im(e^{ix})$. This is much simpler. – Simply Beautiful Art May 27 '17 at 13:05

score 2 · Accepted Answer · answered May 27 '17 at 13:03

Also, you can multiply both sides by $2\sin x$ and apply telescoping:

$$2\sin x\sin x +2\sin 3x\sin x +…+ 2\sin (2n-1)x\sin x=$$ $$[\cos0-\require{cancel}\cancel{\cos2x}]+[\cancel{\cos2x}-\cancel{\cos4x}]+\cdots+[\cancel{\cos(2n-2)x}-\cos2nx]=$$ $$1-\cos2nx$$

Prove by induction that $ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$

2 Answers2