Use the formula
sinθ=1/2i(e^(iθ)-e^(-iθ))
To obtain the identity
sin^5θ=1/16(sin5θ−5sin3θ +10sinθ)
+893 To save on typing, let e^(i.theta) = z.
Then, using the binomial expansion,
$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$
so,
$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$
+118608 I have not obtained the identity yet .
I did check your formula. It seems to me that the 2i needs to be in brackets. 
+893 To save on typing, let e^(i.theta) = z.
Then, using the binomial expansion,
$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$
so,
$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$
+118608 Thanks Bertie,
Thanks anon for asking this great question :))
I took about another 10 steps to get to the end but that is quite neat.
(I used yours as a guide.)
Thank you
Oh Bertie,
I saved some of my birthday cake for you. I shall get it from the fridge
Chris and the regular foum users cooked it for me and it is delicious.
Also it is not as fattening as it looks and I can have an infinite number of big slices from the one cake.
It defies the laws of physics, matter is not conserved, it is endlessly reproduced. 
Enjoy :)
