Formula product of sines

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January 14th 2012, 03:39 AM
wnvl

Formula product of sines

Does anyone know how to prove this formula?

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

All suggestions, ideas are welcome.

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.

Other proofs are also welcome.

Thanks!
January 15th 2012, 02:42 AM
Opalg

Re: Formula product of sines

Quote:

Originally Posted by wnvl

Does anyone know how to prove this formula?

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

I don't know of an easy way to do this. One method would be to start with the Chebyshev polynomials $T_n(s)$ and $U_n(s).$ Using these, you can express $\sin nx$ as a polynomial in $\sin x$ , namely

$\sin nx = \begin{cases}(-1)^{(n-1)/2}T_n(\sin x) & \text{for }n\text{ odd,} \\ (-1)^{(n/2)-1}U_{n-1}(\sin x)\cos x & \text{for }n\text{ even.} \end{cases}$

The properties of Chebyshev polynomials that you need here are that, for n odd, $T_n(s)$ is of the form $T_n(s) = 2^{n-1}s^n + \ldots \pm nx,$ a polynomial of degree n with leading coefficient $2^{n-1}$ and constant term 0. For n even, $U_{n-1}(s)$ is of the form $U_{n-1}(s) = 2^{n-1}s^{n-1} + \ldots \pm nx.$ Again, the leading coefficient is $2^{n-1}$ and the constant term is 0.

For fixed x and n, the numbers $\theta_k = (-1)^k\bigl(x + \tfrac kn\pi\bigr)\ (0\leqslant k\leqslant n-1)$ are all solutions of the equation $\sin n\theta = \sin nx.$ It follows that, if n is odd, then the numbers $\sin \theta_k\ (0\leqslant k\leqslant n-1)$ are all solutions of the equation $(-1)^{(n-1)/2}T_n(s) - \sin nx = 0.$ But the product of the roots of an equation of odd degree is the negative of the constant term divided by the coefficient of the leading term. The product of the roots in this case is $\prod_{k=0}^{n-1}(-1)^k\sin\bigl(x+\tfrac kn\pi\bigr).$ Notice that $(n-1)/2$ of those factors have a negative sign, so that product is equal to $(-1)^{(n-1)/2}\prod_{k=0}^{n-1}\sin\bigl(x+\tfrac kn\pi\bigr).$

The constant term in the equation is $-\sin nx$ , and the coefficient of the leading term is $(-1)^{(n-1)/2}2^{n-1}.$ Thus the formula for the product of the roots tells you that $2^{n-1}\prod_{k=0}^{n-1}\sin\bigl(x+\tfrac kn\pi\bigr) = \sin nx.$

In the case where n is even there is an extra complication caused by the $\cos x$ term in the formula for $\sin nx.$ This corresponds to the factor in the product $\prod_{k=0}^{n-1}\sin\bigl(x+\tfrac kn\pi\bigr)$ given by $k = n/2$ . That factor is $\sin\bigl(x+\tfrac 12\pi\bigr) = \cos x.$ After stripping out that factor from both sides of the identity, you can use the same argument as for the case where n is odd.
January 15th 2012, 06:44 AM
wnvl

Re: Formula product of sines

Thank you Opalg for this great solution. In the meantime I got also a nice, more basic solution on a Dutch math forum where I posted the problem.

If you are interested you can find that solution here Wiskundeforum • Bekijk onderwerp - Bewijs voor sin(nx)