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!



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 $16x^2-\dfrac{256x^4}{3}+\dfrac{8192x^6}{45}$

  • $\begingroup$ Thanks for your help but how did (2n) become (8n)? And why did you write (8n) instead of (8x) (I'm referring to the numerator)? $\endgroup$ – Niko H Mar 15 at 0:49
  • $\begingroup$ @NikoH That was a typo. Now fixed it. $\endgroup$ – Key Flex Mar 15 at 0:52
  • $\begingroup$ But when you plug in n = 0 in the equation, wouldn't you get (1/2) - (1/2) (1 * 1/1), which is 0? $\endgroup$ – Niko H Mar 15 at 0:58
  • $\begingroup$ @NikoH Yes, we do get but question asks for the "first three non-zero terms" $\endgroup$ – Key Flex Mar 15 at 1:01
  • $\begingroup$ oh! thank you much! $\endgroup$ – Niko H Mar 15 at 1:03

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