Math Help - identity help

1. identity help

Given $sinx = \frac{e^{ix} - e^{-ix}}{2i}$ and $cosx = \frac{e^{ix}+ e^{-ix}}{2}$

where $i = \sqrt{-1}$

show that $sin^{4}x = \frac{3}{8} -\frac{1}{2}cos2x + \frac{1}{8} cos4x$

I have a solution for this, provided by my teacher, but i cannot seem to understand it.

the solution is

$sin^{4}x =( \frac{e^{ix}-e^{-ix}}{2})^{4}$ what happened to the i, in the denominator?

this expression is than expanded out, and looks like this; which i dont know how to get to

$\frac{1}{16} ( e^{2ix} + 4e^{2ix} +6 - 4e^{-2ix} + e^{-4ix}$

any help appreciated, thank you.

2. Re: identity help

Originally Posted by Tweety
what happened to the i, in the denominator?
$i^4=1$.

this expression is than expanded out, and looks like this; which i dont know how to get to

$\frac{1}{16} ( e^{2ix} + 4e^{2ix} +6 - 4e^{-2ix} + e^{-4ix}$
Use the Binomial Theorem. (The first term should be $e^{4ix}$).

Thank you.

4. Re: identity help

Actually, just tried expanding using the binomial expansion, but couldn't quite do it.

because in the binomial expansion, the first term has to be 1 $(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!}......$
how do i apply this formula to $e^{ix}-e^{-ix}$?

5. Re: identity help

Use for example the Pascal Triangle (see here: Binomial theorem - Wikipedia, the free encyclopedia),

$\sin^4 x=\dfrac{1}{16}(e^{ix}-e^{-ix})^4=\dfrac{1}{16}(e^{4ix}-4e^{3ix}e^{-ix}+6e^{2ix}e^{-2ix}-4e^{ix}e^{-3ix}+e^{-4ix})$

So,

$\sin^4 x=\dfrac{1}{16}(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix})$