Using De Moivre's theorem to find sin^6(x)

[Faiq]

Homework Statement

Can somebody explain to me how the first two steps are performed?

The Attempt at a Solution

I have no idea how to start the question. I tried using an equation for sin^6 x derived by (cos x + i sinx)^6 = cos 6x+isin 6x but the solution becomes way too hard.

 

[pasmith]

If $z = e^{i\theta}$ then $$\cos n\theta = \frac{z^n + z^{-n}}{2} \\<br /> \sin n\theta = \frac{z^n - z^{-n}}{2i}.$$ Hence $$\sin^6 \theta = \frac{(z - z^{-1})^6}{(2i)^6} = -\frac{(z - z^{-1})^6}{64}.$$

 

[Faiq]

If $z = e^{i\theta}$ then $$\cos n\theta = \frac{z^n + z^{-n}}{2} \\<br /> \sin n\theta = \frac{z^n - z^{-n}}{2i}.$$ Hence $$\sin^6 \theta = \frac{(z - z^{-1})^6}{(2i)^6} = -\frac{(z - z^{-1})^6}{64}.$$

Are these equation correct?
$$\sin^n \theta = \frac{(z-z^{-1})^n}{(2i)^n}$$ and $$\cos^n \theta = \frac{(z+z^{-1})^n}{2^n}$$

 

[John Park]

Yes.