# Proof of sum of sinus terms with mathematical induction.

Hello dear community

I have some issues in solving the following problem:

Proposition: If $x$ is a real number not divisible by $\pi$ ($x\notin \pi\mathbb{Z}$), then for all integers $n \ge 1$ \begin{align} \sin(x) + \sin(3x) + \ldots + \sin((2n-1)x) \\ &= \frac{1-\cos(2nx)}{2\sin(x)} \end{align}

Now this equation should be proved with mathematical induction. I know how to prove simple summations and products (In just showing that $P(k+1)$ follows from $P(k)$ where P is the property to prove (such as the equation in this case).

But in this case, I can't even go beyond the first step in the induction proof. Let me write what I mean, in showing my incomplete proof so far:

Proof (by mathematical induction)

Let the property $P(n)$ be the equation \begin{align} \sin(x) + \sin(3x) + \ldots + \sin((2n-1)x) \\ &= \frac{1-\cos(2nx)}{2\sin(x)} \end{align} Step 1 (Basis Step):
We need to show that $P(1)$ is true. To establish $P(1)$ we must show that the left-hand side of the equation equals to the same quantity as the right-hand side of $P(1)$. The left-hand side is the sum from $\sin(x)$ to $\sin((2\cdot1-1)x)$ which is just $\sin(x)$. The right-hand side is \begin{align} \frac{1 - \cos(2\cdot 1\cdot x)}{2\cdot \sin(x)} = \frac{1 - \cos(2x)}{2\sin(x)} \end{align}

[...] [My proof ends here]

Now I found some connection between the cosinus and sinus function. I think I should use them. $\sin(\alpha) = \cos(\alpha - \pi/2)$ or $\sin^2(\alpha) + \cos^2(\alpha) = 1$

But how should I actually show that $P(1)$ is true? I can't find the appropriate steps, nor do I know how detailed I must depict them, although I know that $P(1)$ must be true. I think the numerator must hold this, in order to fulfill the equation: $1 - \cos(2x) = 2\sin^2(x)$

How would you construct the complete proof? Thanks for any hints and ideas.

Edit:

Since (1) $$\sin^2(\alpha) + \cos^2(\alpha) = 1$$ and (2) $$\cos 2x = \cos^2x-\sin^2x$$ it follows $$\frac{1-\cos(2x)}{2\sin(x)} = \frac{\sin^2(x) + \cos^2(x)-\left(\cos^2(x)-\sin^2(x)\right)}{\sin(x)} = \frac{2\sin^2(x)}{\sin(x)} = \sin(x)$$

Edit No. 2:

To complete the proof:

[...]

Step 2 (inductive step):

We have to show that for all integers $k \ge 1$, if $P(k)$ is true then $P(k+1)$ is true.

Assume that k is any integer with $k \ge 1$ such that $$\sin(x) + \sin(3x) + \ldots + \sin((2k-1)x) = \frac{1-\cos(2kx)}{2\sin(x)}$$ We must show that $P(k+1)$ is true: $$\sin(x) + \sin(3x) + \ldots + \sin((2(k+1)-1)x) = \frac{1-\cos(2(k+1)x)}{2\sin(x)} = \frac{1-\cos(2kx+2x)}{2\sin(x)}$$

But the left-hand side of $P(k+1)$ is: \begin{align} \sin(x) + \sin(3x) + \ldots + \sin((2(k+1)-1)x) \\ & = \Bigg(\sin(x) + \sin(3x) + \ldots + \sin((2k-1)x)\Bigg) + \sin((2k + 1)x) \\ & = \frac{1-\cos(2kx)}{2\sin(x)} + \sin((2k + 1)x) \\ & = \frac{1-\cos(2kx)+\cos(-2kx)+2\sin(x)\sin(2kx+x)}{2\sin(x)} \\ \end{align}

In the next-to-last step we substituted with the inductive hypothesis. Now because of the trigonometric relation $$\sin(x)\sin(y)=\frac12\left(\cos(x-y)-\cos(x+y)\right)$$ we obtain \begin{align} &= \frac{1-\cos(2kx)+\frac12\left(\cos(x-(2kx+x))-\cos(x+2k+x)\right)}{2\sin(x)} \\ &= \frac{1-\cos(2kx)+\cos(-2kx)-\cos(2kx+2x)}{2\sin(x)} \\ \end{align} and because of another trigonometric property $$\cos(x)-\cos(y)=2\sin(\frac{y+x}{2})\sin(\frac{y-x}{2})$$ we receive \begin{align} &= \frac{1-\cos(2kx+2x)+\left(2\sin\left(\frac{2kx-2kx}{2}\right)\sin\left(\frac{2kx--2kx}{2}\right)\right)}{2\sin(x)} \\ &= \frac{1-\cos(2kx+2x)+\left(2\sin(0)\sin(2kx)\right)}{2\sin(x)} \\ &= \frac{1-\cos(2kx+2x)}{2\sin(x)} \\ \end{align}

The last step holds because $\sin(0) = 0$ and the whole product hence becomes zero. And thus we see that the two sides of P(k+1) are equal to the same quantity and therefore they are equal to each other. Thus P(k+1) is true. q.e.d.

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

• Many thanks. (as wel to AlexR). Well, I guess I missed this property. Now I can show that $P(1)$ holds: Since (1) $\sin^2(\alpha) + \cos^2(\alpha) = 1$ and (2) $\cos 2x = \cos^2x-\sin^2x$ it follows $\frac{1 - \cos(2x)}{2\sin(x)} = \frac{\sin^2(x) + \cos^2(x) - \left(\cos^2(x) - \sin^2(x)\right)}{\sin(x)} = \frac{2\sin^2(x)}{\sin(x) = \sin(x)$ Aug 6 '13 at 17:32
• that's right :)
– user71816
Aug 6 '13 at 17:44
• Waah, I guess this crippled latex formula stays here forever :/ See original post for the solution... Aug 6 '13 at 17:45
• you can correct it, change '\cos(2x)' for '\cos (2x)'
– user71816
Aug 6 '13 at 17:50
• No edits allowed after 5 minutes. I'll leave it here. When I delete the post, it will imply just more confusion why we have been commenting right now... Aug 6 '13 at 17:55

Hint Also use the trigonometric Identities (found here):

$$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$ $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$