The initial function is $$h(x)=\arcsin x + \arccos x$$ The derivative of this function is $0$ since $$h'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\equiv0$$

This means that $h(x)$ is a constant function; how can I find the value of $h(x)$?

Could anyone please explain?


3 Answers 3


Try putting $x=0$ - since the function is constant, any convenient value will do.

Cautionary note: beware the domains of definition of the inverse trigonometric functions.

  • $\begingroup$ I'll add that this is the same as one of the basic steps in solving a differential equation--you essentially know $h(x)$ from the integral of $h'(x)$; we can only figure out the "initial condition" by "common sense" $\endgroup$ Jun 23, 2015 at 14:31

@MarkBennet gives the easiest (generally) way to figure out the value of a function once you know it's constant.

I'll only ellaborate here to give an intuitive reason for why this particular function is constant, which doesn't rely on any calculus:

$\arcsin x$ is the measure of the angle whose sine is $x$; $\arccos x$ is the measure of the angle whose cosine is $x$. Let's draw a picture to illustrate the first:


The measure of this angle is $\theta$, but more importantly, notice that this picture also includes $\arccos x$--the cosine of the other acute angle is also $x$.

But we know that the measure of the other angle is just $\frac{\pi}{2}-\theta$, so when we add the two measures together, we get

$$h(x)=\arcsin x + \arccos x=\theta+\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}$$

(basically, no matter what $\theta$ is, it cancels out)

(i.e., at core, this problem is just a fancy way of saying: in a right triangle, the two acute angles sum to $90^\circ$)


Note that $\cos(x)=\sin(\frac{\pi}{2}-x)$. Thus taking $x=\arccos(y)$ and taking the arcsine, we get $\arcsin(y)=\frac{\pi}{2}-\arccos(y)$, thus $$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$


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