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

$\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}$

We have by the double angle identity

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2$

So

$\sin^2(5x)=\dfrac{1-\cos(10x)}2$

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

$\cos(10x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(10x)^{2n}=\sum_{n=0}^\infty\frac{(-100)^n}{(2n)!}x^{2n}$

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