I am trying to find the Taylor series of $\sin^2(4x)$ but I kept getting it wrong. The following is my work:
Apply trig identity
$$\sin^2(4x) = \frac{1-\cos(8x)}{2} $$
Use basic Taylor series which is
$$\cos(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}$$
Plug the Taylor series to the trig function which
$$\sin^2(4x) = \frac12 -\frac12 \sum_{k=0}^\infty (-1)^k \frac{(8x)^{2k}}{(2k)!}$$
Next, I have to find the first three non-zero term. For the first one, all I did was plug $0$ for $k$ in the function and got $0$ but got it wrong. Can anyone help me on this problem? I would appreciate it a lot!
Best Answer
$$\sin^2(4x)$$
We know that $\sin^2(4x)=\dfrac{1-\cos8x}{2}$ and $\cos x=\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}$
$$\cos8x=\sum_{n=0}^{\infty}(-1)^n\dfrac{(8n)^{2n}}{(8n)!}$$ and we get $$\sin^2(4x)=\dfrac{1-\cos8x}{2}=\dfrac12-\dfrac12\sum_{n=0}^{\infty}(-1)^n\dfrac{(8x)^{2n}}{(2n)!}$$
The first non-zero terms are $16x^2-\dfrac{256x^4}{3}+\dfrac{8192x^6}{45}$