Sum of Arcsine and Arccosine

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:

$\exists x \in \closedint {-1} 1: x = \map \cos {y + \dfrac \pi 2}$

Then:

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

Then we can write:

$-y = \arcsin x$

But then:

$\map \cos {y + \dfrac \pi 2} = x$

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

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

$\map {D_x} {\arcsin x + \arccos x} = \dfrac 1 {\sqrt {1 - x^2} } + \dfrac {-1} {\sqrt {1 - x^2}} = 0$

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

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.74$
• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.5 \ (3)$