How to prove $2\arccos(x)+\arccos(1-2x^2)=π$ on $x\in[0,1]$ from MVT First what I did was use the cosine addition formula:
$$2\arccos(x)+\arccos(1-2x^2)=π$$
$$\cos(2\arccos(x))=\cos(π-\arccos(1-2x^2))$$
$$2x^2-1=-(1-2x^2)$$ 
$$0=0$$
However, this is inconsistent with the bound given. Is there anyway I could prove this with the intermediate value theorem? What I first did was let $f(x)=2\arccos(x)+\arccos(1-2x^2)$, and thus as $f(x)$ continuous on $[0,1]$ and differentiable on $(0,1)$ I can use it here. Thus, if $f(x)$ is constant on $[a,b]$ then from the MVT, I get 2 relations:
$$f(0)=f(1)=\pi$$
$$f'(x)=0, 0<x<1$$
The first statement is true but how do I prove the second one? I get the derivative as: 
$$f'(x)=\frac{4 x}{\sqrt{1-\left(1-2 x^2\right)^2}}+\frac{2}{\sqrt{1-x^2}}$$
 A: If 
$$f(x)=2\arccos(x)+\arccos(1-2x^2)$$
Then 
$$\begin{align}
f'(x)&=\frac{-2}{\sqrt{1-x^2}}-\frac{(-4x)}{\sqrt{1-(1-2x^2)^2}}\\
&=\frac{-2}{\sqrt{1-x^2}}+\frac{(2\cdot \sqrt{4x^2})}{\sqrt{4x^2-4x^4}}\\
&=\frac{-2}{\sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}\\
&=0
\end{align}$$
Hence, $f(x)$ is constant. Plug in any point, such as $x=1$, to obtain $$f(x)=2\arccos(x)+\arccos(1-2x^2)=\pi$$
A: Hint: $$\sqrt{1-(1-2x^2)^2}=\sqrt{2x^2}\sqrt{2-2x^2} $$
A: First, recall that $$\cos(-2\theta+\pi)=-\cos(-2\theta)=-\cos(2\theta).$$
On the formula above I've used the fundamental 'summation' formula for the cosine function and a parity argument as well.
By means of the duplication formula 
$$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$$ and of the fundamental trigonometric relation 
$$ \cos^2(\theta)+\sin^2(\theta)=1$$
there holds that
$$\cos(-2\theta+\pi)=1-2\cos^2(\theta).$$
Next, by employing the substitution $\theta=\arccos(x)$ we end up with
$\cos(-2\arccos(x)+\pi)=1-2x^2$
so that $$-2\arccos(x)+\pi=\arccos(1-2x^2)$$
whenever $0\leq -2\arccos(x)+\pi\leq \pi$.
That proves the aforementioned formula.
BTW. I've symply make a proof using fundamental trigonometric relations. In any case you can use the fundamental theorem of calculus to rewrite $2\arccos(x)$ and $\arccos(1-2x^2)$ as the following indefinite integrals
$$
2\arccos(x)-\pi=\int_{0}^{x}\frac{-2}{\sqrt{1-t^2}}~dt.
$$
$$
\arccos(1-2x^2)=\int_{0}^{x}\frac{4u}{\sqrt{1-(1-2u^2)^2}}~du.
$$
Thus, the desired identity may be easily derived by the change of variable $t=1-2u^2$ on the first integral. 
