Question

Prove that `sin^5theta=1/16[sin5theta-5sin3theta+10sintheta]`

Answer 1

Let x = cos 𝜽+𝒊𝒔𝒊𝒏 𝜽  `1/x=costheta-isintheta`

`2costheta=x+1/x`    `sintheta=1/(2i)(x-1/x)`

For sinθ take fifth power on both sides,

`sin^5theta=[1/(2i)(x-1/x)]^5=1/(32i)[x^5-1/x^5-5(x^3-1/x^3)+10(x^1-1/x^1)]`

But `x^n=cosntheta+isinntheta` ,   `x^(-n)=cosntheta-isinntheta`

`x^n-x6(-n)=2isinntheta`

`therefore sin^5theta=1/(32i)[2isinn5theta-5xx2isin3theta+10xx2isintheta]`

`therefore sin^5theta=1/16[sin5theta-5sin3theta+10sintheta]`