I was trying to check whether $ arc sin (sin \theta) = \theta$ and $arc cos(cos \theta) = \theta $ is satisfied when I compose Taylor series expansion of these functions, i.e.:
- For sine:
$$ x = sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ... $$ $$ arc sin x = x + \frac{1}{2 \cdot 3} x^3 + \frac{3}{2 \cdot 4 \cdot 5} x^5 + ... $$
- For cosine:
$$ x = cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - ... $$ $$ arc cos x = \frac{\pi}{2} - (x + \frac{1}{2 \cdot 3} x^3 + \frac{3}{2 \cdot 4 \cdot 5} x^5 + ...) $$
What I wanted to do was to substitute $x$ in the Taylor expansion of an inverse trigonometric function with the Taylor expansion for the respective trigonometric function, group the terms according to the power of $\theta$ and check if everything different than $\theta$ term disappears.
In case of the sine function, I indeed get $\theta = arc sin (sin \theta)$ with this approach. However, in the case of cosine it seems to me that the lowest non-constant term has to be of degree 2. But then it is impossible to get $\theta$ after grouping the terms. Why is this the case? Why does this approach work for sine and does not work for cosine? Am I making a computational mistake in my reasoning or is it somehow connected to the radius of convergence for the series?
$\cos$ is an even function, so $g(\cos(x))$ is an even function for any function $g$. That is, $\cos(-x) = \cos(x)$, so $g(\cos(-x)) = g(\cos(x))$. Thus it is impossible to have $g(\cos(x)) = x$ in a neighbourhood of $0$.