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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: $$f'(x)=\frac{4 x}{\sqrt{1-\left(1-2 x^2\right)^2}}+\frac{2}{\sqrt{1-x^2}}$$

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3 Answers 3

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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 $$f(x)=2\arccos(x)+\arccos(1-2x^2)=\pi$$

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Hint: $$\sqrt{1-(1-2x^2)^2}=\sqrt{2x^2}\sqrt{2-2x^2} $$

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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.

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