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I know that the best way is using infinitesimal calculus but this way we use on the lecture and then we must prove it in the other way which don't use infinitesimal calculus.

My try:
$\lim_{ x\to 1^{-} }\arcsin(x))+\arccos(x))=\lim_{ x\to -1^{+} }\arcsin(x)+\arccos(x)= \frac{\pi}{2} $
So limits of values for extreme $ x $ from the domain of the function are the same.
Moreover sum of continuous functions is continuous functions so $f(x)$ fulfills this property.
That's why I think it is proof that $f(x)=\frac{\pi}{2}$.

However I am afraid that is insufficient. Can you rate this?

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    $\begingroup$ Clearly this is insufficient. Like $g(x)=1-x^2$, you have $g(-1)=g(1)=0$ and $g$ is continuous, but obviously $g$ is no way a constant function. $\endgroup$ – xbh Mar 4 '19 at 12:39
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Notice that $\arcsin y$ is the solution of $y=\sin x$ over $x\in [-\frac{\pi}2,\frac{\pi}2]$ and $\arccos y$ is the solution of $y=\cos x$ over $x\in [0,\pi]$. So if we have $x_0=\arcsin y$, then we have $$ y=\sin x_0 \implies y=\cos\left(\frac \pi 2-x_0\right) $$ and $\frac \pi 2-x_0\in [0,\pi]$. This gives $\arccos y = \frac \pi 2 -x_0=\frac\pi 2 - \arcsin y$ as wanted.

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We have \begin{align} \cos f(x) &= \cos(\overbrace{\arcsin(x)}^{\in\left[-\frac\pi2, \frac\pi2\right]})\cos(\arccos(x)) - \sin(\arcsin(x))\sin(\overbrace{\arccos(x)}^{\in[0,\pi]})\\ &= \sqrt{1-\sin^2(\arcsin(x))}\cos(\arccos(x)) - \sin(\arcsin(x))\sqrt{1-\cos^2(\arccos(x))}\\ &= \sqrt{1-x^2}\cdot x - x \sqrt{1-x^2}\\ &= 0 \end{align} so the image of $f : [-1,1] \to \mathbb{R}$ is contained within $\left\{\frac\pi2 + k\pi : k \in \mathbb{Z}\right\}$. Since $f$ is continuous and $[-1,1]$ is connected, we conclude that image of $f$ must be a singleton so $f$ is constant.

Plugging in $x = 0$ yields $f \equiv \frac\pi2$.

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Notice that $f'(x) = 0$ for all $x \in \mathbb{R}$. So $f(x) = c$ for a constant $c \in \mathbb{R}.$ Since $f(0) = \frac{\pi}{2}$, one has $c = \frac{\pi}{2}.$ Hence the conclusion.

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