Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$

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: $$\begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array}$$

• Have you tried drawing a triangle? – GFauxPas Jan 23 '15 at 21:26
• Yes but got confused... Also tried viewing their graphs using Google search - and I can see why this is true. – Dor Jan 23 '15 at 21:28

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?

• Bleugh. Sorry about the weird wording at first. Not having a diagram at your disposal makes it hard to tell what is going on exactly. The way to show that this works for any $x$ in that range is to set your triangle up in such a way that $\frac{b}{c} = x$. For instance, take $b = x$, $c=1$ and $a = \sqrt{1-x^2}$ with appropriate signs. – Cameron Williams Jan 23 '15 at 21:36
• Also.. The calculus based argument is in some sense a repeat of the original argument since the first observation is often used to determine what $\arccos(x)'$ is. – Cameron Williams Jan 23 '15 at 21:37
• Could you please elaborate regarding the calculus based argument? Using the limits of integration (similar to what I've written in my first post).. It would really help in my understanding of calculus..thanks! – Dor Jan 23 '15 at 21:41
• @Dor The problem is that you're choosing two different lower limits of integration. Try picking a specific lower limit of integration, say $0$, instead of $C_1$ and $C_2$. The fundamental theorem basically tells you this is okay. When you do, you'll get your answer. Another reason your approach fails is that integrals are only additive (with regards to the integrand, that is) if the limits of integration match. Yours do not. – Cameron Williams Jan 23 '15 at 21:43
• How does your first argument prove the result when $x$ is negative? – user208259 Jan 23 '15 at 22:06

It is pretty obvious to see that $\sin$ and $\cos$ are the same curve, just shifted by $\pi/2$, so if you consider following craphis, it should be clear:

• Indeed I've seen their graphs.. but why would the integral in my first post is true? – Dor Jan 23 '15 at 21:30

More simple.... From $\cos \alpha=\sin \left(\dfrac{\pi}{2}-\alpha\right)$ we have: $$\cos y=x \Rightarrow \sin\left(\dfrac{\pi}{2}-y\right)=x \Rightarrow$$ $$\Rightarrow \begin{cases} \arccos x=y \\ \arcsin x= \dfrac{\pi}{2}-y \end{cases} \Rightarrow$$ $$\Rightarrow \arccos x+\arcsin x=\dfrac{\pi}{2}$$

By definition, $\arcsin(x)$ is the angle $\alpha$ such that $\sin(\alpha) = x$ and $-\pi/2 \le \alpha \le \pi/2$, while $\arccos(x)$ is the angle $\beta$ such that $\cos(\beta) = x$ and $0 \le \beta \le \pi$. Since $-\pi/2 \le \alpha \le \pi/2$, $\cos(\alpha) \ge 0$, so we have $\cos(\alpha) = \sqrt{1 - x^2}$. Similarly $\sin(\beta) = \sqrt{1-x^2}$. Now \eqalign{-\pi/2 &\le \arcsin(x) + \arccos(x) = \alpha + \beta \le 3 \pi/2 \cr\sin(\alpha + \beta) &= \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) = x^2 + 1 - x^2 = 1\cr \cos(\alpha + \beta) &= \cos(\alpha) \cos(\beta) - \sin(\alpha)\sin(\beta) = \sqrt{1-x^2} x - x \sqrt{1-x^2} = 0} and the only angle in this interval with that sine and cosine is $\pi/2$.

Write $\theta = \arcsin x$. By definition, this means: $$\sin \theta = x, \qquad -\pi/2 \leq \theta \leq \pi/2.$$

You want to show that $\pi/2 - \theta = \arccos x$. By definition, this means: $$\cos(\pi/2 - \theta) = x, \qquad 0 \leq \pi/2 - \theta \leq \pi.$$

The last inequality follows immediately from the bounds for $\theta$. Furthermore, we have $\cos(\pi/2 - \theta) = \sin \theta = x$ by the complementary angle formula.

Note: To prove the complementary angle formula $\cos(\pi/2 - \theta) = \sin \theta$ in general, an argument with a triangle is not enough, since it will only be valid for acute angles. It is preferable to consider the effect of the reflection through the line $y = x$ on points of the unit circle.

In this figure taken from Trigonometrija by Gel'fand, L'vovskij and Toom, the vertex $B = ?$ of the shaded triangle that is on the circle is the reflection through the diagonal (not shown) of the point $A$ with angular coordinate $x$. The angular coordinate of $B$ is $\pi/2 - x$, so $B = (\cos(\pi/2-x),\sin(\pi/2-x))$. We also have $A = (\cos x,\sin x)$. On the other hand, this reflection takes a point with coordinates $(a,b)$ to the one with coordinates $(b,a)$. Thus $B=(\sin x, \cos x)$, proving that $\cos(\pi/2 - x) = \sin x$. This argument is valid even if $x$ isn't in the first quadrant.

You seem to be starting with the observation that $$\frac{1}{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}}=0$$ which has the consequence that, for any $c_1,c_2\in[-1,1]$, $$\int_{c_1}^{x}\frac{1}{\sqrt{1-x^2}}\,dx+\int_{c_2}^{x}\frac{-1}{\sqrt{1-x^2}}\,dx$$ is constant, but not necessarily $0$.

Indeed \begin{multline} \int_{c_1}^{x}\frac{1}{\sqrt{1-x^2}}\,dx+\int_{c_2}^{x}\frac{-1}{\sqrt{1-x^2}}\,dx =\\ \int_{c_1}^{x}\frac{1}{\sqrt{1-x^2}}\,dx+ \int_{c_2}^{c_1}\frac{-1}{\sqrt{1-x^2}}\,dx+ \int_{c_1}^{x}\frac{-1}{\sqrt{1-x^2}}\,dx \end{multline} and this sum is just $$\int_{c_1}^{c_2}\frac{1}{\sqrt{1-x^2}}\,dx=\arcsin c_2-\arcsin c_1$$

So the sum of the integrals you're computing is zero only if $c_1=c_2$.

The fact that $\arcsin x+\arccos x=\pi/2$ follows from differentiating: the function $f(x)=\arcsin x+\arccos x$ has zero derivative on $(-1,1)$, so it's constant in that interval and, being continuous on $[-1,1]$ it is constant also in $[-1,1]$. The constant can be evaluated as $$f(0)=\arcsin 0+\arccos0=0+\frac{\pi}{2}=\frac{\pi}{2}.$$

first let us deal with the the first quadrant. we will pick two points $A = (x,y), B = (y,x)$ on the unit circle in the first quadrant. the two points are images of each other on the mirror along the line $y = x$ let us also label the points $P = (1,0), Q = (0,1)$

we will need the following two things:

(a) $arc PA = arc QB,$

(b) $arc PB + arc PA = \pi/2.$

let us use the definition, you can see why the name is apt,
$\arcsin y = \arccos x = arc PA$ where $A = (x, y)$ is a point on the unit circle.

using the above definition $$\arccos x = arc AP, \arcsin x = arc BP, \arccos x + \arcsin x = arc AP + arc BP = \pi/2$$

the range of $\arccos$ is $[0, \pi]$ and of $\arcsin$ is $[-\pi/2, \pi/2]$. so that when you have negative argument for $arccos$ the point in the second quadrant, and for $\arcsin$ is in the fourth quadrant. the same argument works when $x< 0.$