Proving that $\arcsin(x)+\arccos(x)= 90^{\circ}$

Question

The question starts with $y=\arccos(x)$ and asks to express $\arcsin(x)$ in the form of $y$.

I got $x=\cos(y)$ and then used $\cos(y)=\sin(y+90)$, so $x=\sin(y+90^{\circ})$ and then $\arcsin(x)=y+90^{\circ}$.

From there I then said that if $\arcsin(x)=y+90^{\circ}$ and $\arccos(x)=y$, $\arcsin(x)+\arccos(x)= 2y+90^{\circ}$, but I know it should just be $90^{\circ}$.

The solution uses $\cos(y)=\sin(90^{\circ}-y)$ but since $\sin(90^{\circ}-y)=\sin(y+90^{\circ})$. I don't understand why my answer comes out differently.

$y=\arccos x$ implies $\cos y=x$ but $a=\cos b$ does not imply $b=\arccos a$. — Kavi Rama Murthy, Apr 29 '20 at 12:02
For example $0 =\cos (\frac {\pi} 2+1000\pi)$. But $\frac {\pi} 2+1000\pi \neq \arccos 0$. See the definitioin of $\arccos$ to see why $\arccos $ cannot be large number. In fact $arccos 0=\pi/2$ not $\pi /2 +1000\pi$. — Kavi Rama Murthy, Apr 29 '20 at 12:06
So how would I use this in my answer? How would I know to use sin(90-y) rather than sin(y+90)? — zacccczn, Apr 29 '20 at 12:08
https://math.stackexchange.com/questions/1176711/is-it-generally-true-that-arcsin-theta-arccos-theta-frac-pi2 — lab bhattacharjee, Apr 29 '20 at 12:13

score 2 · Answer 1 · answered Apr 29 '20 at 12:24

One way to do this class of problems without running into issues in boundary of the domains is geometrically. Since you have $y = \arccos(x) \implies x = \cos y$ think about a right triangle $\Delta ABC$ with $\angle B = 90^\circ$ and $y = \angle A$.

You are free to scale this any way you like, so let's pick the hypotenuse $AC=1$ and then your constraint says $AB=x$.

Now you need to find $\arcsin(x)$, in other words, the angle for which $x$ would be the sine. Looking at our $\Delta ABC$ it is easy to see that $\sin C = x$ and the measure of $C$ must be $$ \angle C = 180^\circ - \angle A - \angle B = 180^\circ - y - 90^\circ = 90-y. $$

score 2 · Answer 2 · answered Apr 29 '20 at 12:38

By definition, $y=\arccos x\iff \cos y=x \textbf{ and } y\in[0,\pi]$.

On the other hand, $y=\arcsin x\iff \sin y=x \textbf{ and } y\in\bigl[-\frac \pi 2,\frac\pi 2\bigr]$. Taking into account the intervals condition, you get $$\arcsin x=\frac\pi 2-y.$$

Proving that $\arcsin(x)+\arccos(x)= 90^{\circ}$

2 Answers2