MHB What Are the First Two Terms of sin(sin(2x))?

[Guest2]

I'm asked to find first two terms of the series $\sin(\sin(2x))$.

$\sin(t) = t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots$

$\sin(2x) = 2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots$

$\displaystyle \sin(\sin(2x)) = (2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots)-\frac{(2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots)^3
}{3!}+\cdots$

The cubed term and those beyond don't contribute anything to the first two terms, so

So $\sin(\sin(2x)) = 2x-\frac{2^3x^3}{3!}+\cdots$

However, this is wrong. What have I missed?

 

[Prove It]

I'm asked to find first two terms of the series $\sin(\sin(2x))$.

$\sin(t) = t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots$

$\sin(2x) = 2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots$

$\displaystyle \sin(\sin(2x)) = (2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots)-\frac{(2x-\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\cdots)^3
}{3!}+\cdots$

The cubed term and those beyond don't contribute anything to the first two terms, so

So $\sin(\sin(2x)) = 2x-\frac{2^3x^3}{3!}+\cdots$

However, this is wrong. What have I missed?

You missed an $\displaystyle \begin{align*} x^3 \end{align*}$ term...

 

[Guest2]

You missed an $\displaystyle \begin{align*} x^3 \end{align*}$ term...

You're right, thanks. So the answer is $\displaystyle \sin(\sin(2x)) = 2x - \frac{8x^3}{3} + \mathcal{O}{(x^4)}.$