The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\
\arcsin(x) – \arcsin(C1) + \arccos(x) – \arccos(C2) = 0 \\
\text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$
I can't find the right words to explain why this is true?
Edit #1 (25 Jan, 20:10 UTC):
The following is a truth clause:
$$
\begin{array}{ll}
\frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\
\\
\frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0
\end{array}
$$
By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) – \arcsin(k) + \arccos(x) – \arccos(k) = m \text{ (m is a constant)}\\
\\
\arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\
\\
\text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Using Calculus, why is that true for every $x \in [-1,1]$?
Edit #2:
A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.
Thus the equations from "Edit #1" should be as follows:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) – \arcsin(k) + \arccos(x) – \arccos(k) = 0\\
\\
\arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\
\\
A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Best Answer
There are a couple of ways to see this. Firstly, draw a right triangle, call it $ABC$ (with $C$ being the right angle), with side lengths $a$, $b$ and $c$ with the usual convention. Then $\arcsin(\frac{b}{c})$ is the measure of the angle $CBA$. Additionally, $\arccos(\frac{b}{c})$ is the angle of the angle of the opposite angle $CAB$, so $\arccos(\frac{b}{c}) = \frac{\pi}{2}-\arcsin(\frac{b}{c})$ since the opposite angles must sum to $\frac{\pi}{2}$. From here, you get the result.
We could also do some calculus to figure it out. Let's let $f(x) = \arcsin(x)+\arccos(x)$. Then $f'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$. Thus $f$ is constant. What is $f(0)$ equal to?