How do I use this fact to find $\sin^5\theta$?

Question

I have a complex number, $z,$ which has a modulus of $1$ and argument of $\theta$.

The first part is to show that $z^n-\frac{1}{z^n}=2i \sin(n\theta)$, which I have done.

The next question is this: Hence show that $\sin^5\theta = \frac{1}{16}(\sin 5\theta - 5\sin3\theta + 10\sin\theta)$

How would I use the first part to show the second?

Answer 1

Hint:

Put $n=5,3,1$ in $$z^n-\dfrac1{z^n}=2i\sin(n\theta)$$

and calculate $$\sin5\theta-5\sin3\theta+10\sin\theta$$

Answer 2

Hint

Use $$\left(z-\dfrac 1z\right)^5=z^5-\dfrac 1{z^5}-\binom51\left (z^3-\dfrac 1{z^3}\right)+\binom 52\left (z -\dfrac 1z\right)$$ as $$\binom nr =\binom n{n-r}$$

Now put $n=1,3,5$