Question

How do you find the Maclaurin Series for `sin^3 (x)`?

Answer 1

`sin^3 x = sum_(n=0)^oo (((-1)^n3(1-3^(2n)) )/(4(2n+1)!)x^(2n+1))`

`= x^3-x^5/2+(13x^7)/120-(41x^9)/3024+...`

Explanation:

It is probably easiest to derive it from the Maclaurin series for `sin x` and the triple angle formula for `sin`:

`sin x = sum_(n=0)^oo ((-1)^n )/((2n+1)!)x^(2n+1) = x - x^3/(3!) + x^5/(5!) - x^7/(7!) +...`

`sin 3 x = 3 sin x - 4 sin^3 x`

So:

`sin^3 x = 1/4 (3 sin x - sin 3 x)`

`color(white)(sin^3 x) = 1/4 (3 (sum_(n=0)^oo ((-1)^n x^(2n+1))/((2n+1)!)) - (sum_(n=0)^oo ((-1)^n (3x)^(2n+1))/((2n+1)!)))`

`color(white)(sin^3 x) = 1/4 sum_(n=0)^oo (3((-1)^n x^(2n+1))/((2n+1)!)- ((-1)^n (3x)^(2n+1))/((2n+1)!))`

`color(white)(sin^3 x) = 1/4 sum_(n=0)^oo (((-1)^n(3-3^(2n+1)) x^(2n+1))/((2n+1)!))`

`color(white)(sin^3 x) = sum_(n=0)^oo (((-1)^n3(1-3^(2n)) )/(4(2n+1)!)x^(2n+1))`