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