I'm working from the text "Intro To Real Analysis" by William Trench. Here is what I have thus far.
I will prove using Mathematical Induction that $\sin x+\sin 3x+…+\sin (2n-1)x=\frac{1-\cos 2nx}{2\sin x}, $ where $n\ge 1.$ Consider the case where $n=1$, which will serve as our base case. Clearly, for $n=1$ it is true that $$\sin x = \frac{1-\cos 2x}{2\sin x}$$ since $\sin^2 x = \frac{1-\cos 2x}{2}$ is an identity. Next, we assume that $\sin x+\sin 3x+…+\sin (2n-1)x=\frac{1-\cos 2nx}{2\sin x} $ is true for $n$ and then using Mathematical Induction, we verify the $n+1$ case, that is to say, we observe that $$\frac{1-\cos 2(n+1)x}{2\sin x} = \frac{1 – \cos(2nx+2x)}{2\sin x}$$ $$ = \frac{1-\cos(2nx)\cos(2x)-\sin(2nx)\sin(2x)}{2\sin x}$$ $$ =???.$$
I'm not sure where I can go from here. I realize I want to transform the last line into $\sin x+\sin 3x+…+\sin (2n)x$, but can't see how I can get there.
Any helpful hints would be great! Thanks!
Best Answer
Hint:
$$\cos 2nx\cos 2n+\sin 2nx\sin 2x=\cos(2nx-2x)=\cos\left(2x(n-1)\right)$$
Edited:
$$\sin x+\sin 3x\ldots+\sin((2n-1)x+\sin((2n+1)x\stackrel{\text{Ind. Hypothesis}}=$$
$$=\frac{1-\cos2nx}{2\sin x}+\sin((2n+1)x$$
Observe that the added term for $\;n+1\;$ is $\;\sin((2n+1)x\;$ , not $\;\sin 2nx\;$ !
Now, you want the last line above to equal
$$\frac{1-\cos2nx}{2\sin x}+\sin((2n+1)x)=\frac{\cos((2n+2)x)}{2\sin x}\iff$$
$$\frac{\cos((2n+2)x)+\cos2nx-1}{2\sin x}=\sin((2n+1)x)\iff$$
$$\cos 2nx\cos2x-\sin2nx\sin2x+\cos2n x=2\sin x\sin((2n+1)x)$$
Take it from here now.