Taylor development of $\arctan(\cos(x))$ near $0$

Question

How would I find the "Taylor development of $\arctan(cos(x))$ near $0$ at order $5$?"

I am translating that from french, so I am not sure how I have to call it it english.

By order $5$ I mean that I want to find $\arctan(\cos(x))=\sum_{k=0}^{5}$TaylorFormulaHere $+ o(x^5)$.

Answer 1

Hint: Let $u(x)=\cos(x)=\left(1-\displaystyle\frac{x^2}{2!}+\frac{x^4}{4!}+\dots\right)$.
Write up also the first some terms of the Taylor series $\arctan(u)$, using variable $u$ in place of $x$.
Then substitute.

Answer 2

Hint

Let $f(x)=\arctan(\cos x)$ then $$f'(x)=-\frac{\sin x}{1+\cos^2 x}$$ so write the Taylor's series of $f'(x)$ using the Taylor series of $\sin x$ and $\cos x$ and $\frac1{1+x}$ and then we find the Taylor's series of $f$ by $$f(x)=\int_0^xf'(t)dt+\frac\pi4$$

Answer 3

I assume that the "Taylor Development" is the Taylor Series. We have, close to $x=p$, $$\mathrm{f}(x) \sim \mathrm{f}(p) + \mathrm{f}'(p)\cdot (x-p) + \frac{\mathrm{f}''(p)}{2!}\cdot (x-p)^2 + \cdots + \frac{\mathrm{f}^{(k)}(p)}{k!}\cdot (x-p)^k + \cdots$$

In our case, we have $x=0$. We need to find the values of the first five derivatives at $x=0$.

If $\mathrm{f}(x) = \arctan\left(\cos x\right)$, then $\mathrm{f}(0) = \frac{\pi}{4}$. For the derivatives, we can apply the chain rule.

The key fact that we need to know is:

$$\frac{\mathrm{d}}{\mathrm{d}y} \arctan y = \frac{1}{1+y^2}$$

We can apply the chain rule to find the derivative $\mathbb{f}'(x)$. We have:

$$\mathrm{f}'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\arctan\left(\cos x \right) = \frac{1}{1+\cos^2x}\cdot(-\sin x) \equiv -\frac{\sin x}{1+\cos^2x}$$

It follows that $\mathrm{f}'(0) = 0$. Next, we need to find the second derivative $\mathrm{f}''(x)$. We can use the quotient rule to do this, and then use trigonometric identities to simplify when possible:

$$\mathrm{f}''(x) = \frac{\mathrm{d}^2}{\mathrm{d}x^2}\arctan\left(\cos x \right) =\frac{\mathrm{d}}{\mathrm{d}x}\!\!\left(-\frac{\sin x}{1+\cos^2x}\right)=\frac{\left(\cos^2x - 3\right)\cos x}{\left(1+\cos^2x\right)^2}$$ It follows that $\mathrm{f}''(0) = -\frac{1}{2}$. So far, this tells us that $$\arctan\left(\cos x\right) \sim \frac{\pi}{4} - \frac{1}{4}x^2 + \cdots$$

To find the rest of the Taylor Series, you need to find the next three derivatives.