Question

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(5x) based at 0.

Answer 1

Presumably you know the Taylor series for
`\cos x` to be

`\cos x=\displaystyle\sum_(n=0)^\infty((-1)^n)/((2n)!)x^(2n)`

We have by the double angle identity

`\sin^2\theta=\frac{1-\cos(2\theta)}2`

So

`\sin^2(5x)=\frac{1-\cos(10x)}2`

and substituting
`10x` for
`x` in the Taylor series above gives

`\cos(10x)=\displaystyle\sum_(n=0)^\infty((-1)^n)/((2n)!)(10x)^(2n)=\sum_(n=0)^\infty((-100)^n)/((2n)!)x^(2n)`

`\implies\boxed{\sin^2(5x)=\displaystyle\frac12-\frac12\sum_(n=0)^\infty((-100)^n)/((2n)!)x^(2n)}`