Sum of Arcsine and Arccosine

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Theorem

Let $x \in \R$ be a real number such that $-1 \le x \le 1$.


Then:

$\arcsin x + \arccos x = \dfrac \pi 2$

where $\arcsin$ and $\arccos$ denote arcsine and arccosine respectively.


Proof

Let $y \in \R$ such that:

$\exists x \in \left[{-1 \,.\,.\, 1}\right]: x = \cos \left({y + \dfrac \pi 2}\right)$

Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x\) \(=\) \(\displaystyle \) \(\displaystyle \cos \left({y + \frac \pi 2}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle -\sin y\) \(\displaystyle \) \(\displaystyle \)          Cosine of Angle plus Right Angle          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sin \left({-y}\right)\) \(\displaystyle \) \(\displaystyle \)          Sine Function is Odd          


Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.

Then we can write $-y = \arcsin x$.

But then $\cos \left({y + \dfrac \pi 2}\right) = x$.

Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.

Hence $y + \dfrac \pi 2 = \arccos x$.

That is, $\dfrac \pi 2 = \arccos x + \arcsin x$.

$\blacksquare$


Note

Note that from Derivative of Arcsine Function and Derivative of Arccosine Function, we have:

$D_x \left({\arcsin x + \arccos x}\right) = \dfrac 1 {\sqrt {1 - x^2}} + \dfrac {-1} {\sqrt {1 - x^2}} = 0$

which is what (from Derivative of Constant) we would expect.


Sources