Using induction to prove a formula for $\sin x+\sin 3x+\dots+\sin (2n-1)x$

Question

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!

Answer 1

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.

Answer 2

The induction hypothesis is $$ \sin x+\sin 3x+\dots+\sin (2n-1)x=\frac{1-\cos 2nx}{2\sin x} $$ so $$ \sin x+\sin 3x+\dots+\sin (2n-1)x+\sin(2n+1)x= \frac{1-\cos 2nx}{2\sin x}+\sin(2n+1)x $$ and the right hand side can be written as $$ \frac{1-\cos2nx+2\sin(2n+1)x\sin x}{2\sin x} $$ so we are bound to prove that $$ 1-\cos2nx+2\sin(2n+1)x\sin x=1-\cos2(n+1)x $$ or $$ \cos2(n+1)x-\cos2nx=-2\sin(2n+1)x\sin x $$ By prostapheresis, $$ \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} $$ and indeed, $$ \frac{2(n+1)x+2nx}{2}=(2n+1)x,\qquad \frac{2(n+1)x-2nx}{2}=x. $$

Without knowing prostapheresis, you can write $$ 2(n+1)x=(2n+1)x+x $$ and $$ 2nx=(2n+1)x-x $$ and apply the addition and subtraction formulas for the cosine.