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I want to prove the following by induction $$\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)$$ here's my trial:

Base Step

N = 1 we have that $LHS = \sin(x)\cos(x)$ and $RHS = \frac{1}{2}\sin(2x) = \sin(x)\cos(x)$. Hence it works for N=1.

Inductive Step

Suppose it works for some $N$, i.e. $$\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)$$ I want to show that this implies that $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2(N+1)x)$$

My attempt

$$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) + \sin(x)\cos((2(N+1)-1)x)$$ which gives $$ \sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)+\sin(x)\cos((2N+1)x)$$

Now taking the last term I can see that $$\sin(x)\cos((2N-1)x) = \sin(x)(\cos(2Nx)\cos(x)-\sin(2Nx)\sin(x))$$ which gives $$\sin(x)\cos((2N-1)x) = \frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$

But I really can't go any further.. any suggestions?

How do I prove it by induction?

EDIT

So in total we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)+\frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$ and rearranging we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sin(2Nx)\left(\frac{1}{2}-\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ and again $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\left(1-2\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so finally $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\cos(2x)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so then it's trivial $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx+2x) = \frac{1}{2}\sin(2(N+1)x)$$

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  • $\begingroup$ Shouldn't it be $\sin(x)\cos((2N+1)x)$ instead of $\sin(x)\cos((2N-1)x)$? $2(N+1)-1=2N+1$. $\endgroup$ – Clement C. Mar 9 '17 at 19:59
  • $\begingroup$ @ClementC. yes sure, that was a typo! $\endgroup$ – Euler_Salter Mar 9 '17 at 20:03
  • $\begingroup$ Okay, I managed to solve it ahah $\endgroup$ – Euler_Salter Mar 9 '17 at 20:10
  • $\begingroup$ You may want to post your solution as an answer and accept it so this question doesn't appear unanswered. $\endgroup$ – Mark S. Mar 30 '17 at 23:15
  • $\begingroup$ @Mark S. I will tomorrow, thank you $\endgroup$ – Euler_Salter Mar 31 '17 at 0:36
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So in total we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)+\frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$ and rearranging we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sin(2Nx)\left(\frac{1}{2}-\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ and again $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\left(1-2\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so finally $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\cos(2x)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so then it's trivial $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx+2x) = \frac{1}{2}\sin(2(N+1)x)$$

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