Sum of Arcsine and Arccosine

Theorem

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

Let:

• $\arcsin x$ be the arcsine of $x$.
• $\arccos x$ be the arccosine of $x$.

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

Proof

Let $y \in \R$.

Then $\cos \left({y + \dfrac \pi 2}\right) = - \sin y = \sin \left({-x}\right)$.

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

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

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

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $5.74$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.5 \ (3)$