0

I'm having trouble proving the below statement using the product to sums and sums to product identities in trigonometry:

We were told to use the the result $$\sin A\sin B = \frac12 \left(\cos(A-B)-\cos(A+B)\;\right)$$ to prove that $$\sin x + \sin 3x + \sin 5x + \cdots + \sin(2n-1)x = \frac{\sin^2 nx}{\sin x}$$

user218115
  • 3

1 Answers1

0

Suppose $\sin x \ne 0$. Then \begin{align*} \sum_{i=1}^n \sin((2i - 1)x) & = \frac{\sin x}{\sin x}\sum_{i=1}^n \sin((2i - 1)x) \\ & = \frac 1{\sin x}\sum_{i=1}^n \sin((2i - 1)x) \sin x. \end{align*} Now use the given formula for each summand on the right-hand side: \begin{align*} \sum_{i=1}^n \sin((2i - 1)x) & = \frac 1{\sin x}\sum_{i=1}^n \frac 12(\cos(2(i - 1)x) - \cos(2ix)). \end{align*} The sum on the right-hand side is telescoping, so \begin{align*} \sum_{i=1}^n \sin((2i - 1)x) & = \frac 1{\sin x}\left(\frac{1 - \cos(2nx)}2\right) \\ & = \frac{\sin^2(nx)}{\sin x}. \end{align*}

Tunococ
  • 10,303
  • Thank you! I just have two more questions; how do we know that the sum on the right hand side is telescoping? And where is the 1-cos(2nx) coming from? – user218115 Feb 21 '15 at 23:45
  • Whenever you have a sum of the form $\sum_{i=1}^n (a_{i-1} - a_i)$, you can rearrange it as $$\left(a_0 + \sum_{i=2}^n a_{i-1}\right) - \left(a_n + \sum_{i=1}^{n-1} a_i\right).$$ Reindex the first summation to $\sum_{i=1}^{n-1} a_i$, and you can see that the two summations cancel, leaving you with $a_0 - a_n$. In this particular problem, $a_i = \cos(2ix)$, so you are left with $a_0 - a_n = \cos(0) - \cos(2nx)$. – Tunococ Feb 22 '15 at 00:22