Proving the Equation with Complex Numbers

[wnvl]

Homework Statement

Prove the following equation

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

Homework Equations

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.

 

[flatmaster]

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

 

[Dick]

Is k an arbitrary integer?

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

 

[wnvl]

In the mean time I got a nice solution on this dutch math forum.


http://www.wiskundeforum.nl/viewtopic.php?f=10&t=5579&p=37185#p37177