# Formula product of sines

1. Jan 14, 2012

### wnvl

1. The problem statement, all variables and given/known data

Prove the following equation

$$2^{n-1}\prod_{k=0}^{n-1}\sin(x+\frac{k\pi}{n}) = \sin(nx)$$

2. Relevant equations

3. The attempt at a solution

Below you find my unsuccessfull attempt using complex numbers.

When you convert it to complex numbers the equality can be rewritten as

$$2^{n-1}\prod_{k=0}^{n-1}\frac{e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})}}{2i} = \frac{e^{inx}-e^{-inx}}{2i}$$
$$i^{-n+1}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}$$
$$e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}$$

The LHS of this equation contains terms in $$e^{inx}$$, $$e^{i(n-1)x}$$, $$e^{i(n-2)x}$$, ..., $$e^{-inx}$$.
Calculate the coefficient for each term.

$$\underline{e^{inx}}$$

$$e^{i\frac{(-n+1)\pi}{2}}\frac{(-n+1)\pi}{2}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}=e^{i\frac{(-n+1)\pi}{2}}e^{inx+i\sum_{k=0}^{n-1}\frac{k\pi}{n}}=e^{inx}$$

$$\underline{e^{-inx}}$$

$$e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}-e^{-i(x+\frac{k\pi}{n})}=(-1)^ne^{i(-n+1)\pi}e^{-inx}=(-1)^{n}e^{-inx} = -e^{-inx}$$

Now we still have to prove that for -n<k<n the coefficient of $${e^{ikx}}$$ equals 0 to conclude the proof. But I don't know how to do this.

All suggestions, ideas are welcome. Proofs not using complex numbers will be appreciated as well.

Last edited: Jan 14, 2012
2. Jan 15, 2012

### flatmaster

Taylor expansion of Sin(x+kpi/n) about the point x=kpi/n ?

3. Jan 15, 2012

### Dick

k is an index of the product. Of course it's not random. It goes from 0 to n-1.

4. Jan 15, 2012