Use the formula

sinθ=1/2i(e^(iθ)-e^(-iθ))

To obtain the identity

sin^5θ=1/16(sin5θ−5sin3θ +10sinθ)

Guest Apr 14, 2015

#3**+5 **

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,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

Bertie Apr 17, 2015

#1**+5 **

I have not obtained the identity yet .

I did check your formula. It seems to me that the 2i needs to be in brackets.

Melody Apr 15, 2015

#3**+5 **

Best Answer

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,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

Bertie Apr 17, 2015

#4**0 **

**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 :)

Melody Apr 17, 2015