Taylor expansion of $\sin(\cos x^2)$ in x=0

Question

I don't understand how to do the series expansion of $$\sin(\cos x^2)$$ in x=0, up to fourth order of derivation. I know the composite function developments, but mixing McLaurin (x=0) and taylor (x!=0) is confusional: I tried to to the expansion of $y=\cos x^2$, but once I do it and replace the "new" y in the expansion of sen(1), I don't think to have a polynomial as a result.

Answer 1

Just start, as you did, with $$\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right)$$ So $$\cos(x^2)=1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)$$ So $$A=\sin(\cos(x^2))=\sin\left(1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)\right)$$ Define $a=\frac{x^4}{2}-\frac{x^8}{24}+\cdots$ which makes $$A=\sin(1-a)=\sin(1)\cos(a)-\sin(a)\cos(1)$$ and use again Taylor with $a$ to get $$A=\sin (1)-a \cos (1)-\frac{1}{2} a^2 \sin (1)+\frac{1}{6} a^3 \cos (1)+O\left(a^4\right)$$ Replace now $a$ by $\frac{x^4}{2}-\frac{x^8}{24}+\cdots$