Inverse trigonometric identities

Inverse trigonometric functions already name one angle on each principal branch. How do those names rewrite each other, and what slope does each graph carry?

Two principal values add to a right angle

This picture keeps $x$ strictly between $-1$ and $1$. Drag it. The green point rides $y=\arcsin x$. The blue point has height $\pi /2-\arccos x$. The segment between them stays level, so the two principal values of one number add to a right angle.

For every real $x$ with $-1<x<1$,

$$\arcsin x=-\arcsin (-x)=\frac{\pi }{2}-\arccos x=\arctan \frac{x}{\sqrt{1-x^2}}.$$

The first equality is oddness on that branch. The level segment is the middle equality. The third reads the same angle as an arctan, with opposite side $x$ and radical $\sqrt{1-x^2}$.

The same sum for arctan

For every real $x$ the extra cut is different. The input is no longer trapped between $-1$ and $1$, and the radical is $\sqrt{1+x^2}$.

$$\arctan x=-\arctan (-x)=\frac{\pi }{2}-\mathrm{arccot}x=\arcsin \frac{x}{\sqrt{1+x^2}}.$$

Oddness is again the first equality. The middle equality is the co-angle with arccot. The third reads that same principal angle as an arcsin.

One step reads the slope

For every real $x$ with $-1<x<1$,

$$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}.$$

For every real $x$,

$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\frac{d}{dx}\mathrm{arccot}x=-\frac{1}{1+x^2}.$$

Arccos and arccot flip the sign of their partners. The arcsin slope equals $1$ at $x=0$ and grows as $x$ approaches either endpoint. The arctan slope stays strictly between $0$ and $1$, and it equals $1$ at $x=0$.

Drag the point on $y=\arcsin x$. The horizontal step has length $1$. The vertical step equals $1/\sqrt{1-x^2}$.

Drag the point on $y=\arctan x$. The horizontal step has length $1$. The vertical step equals $1/(1+x^2)$.

The identities move among the four principal names, and one horizontal step reads the slope on each graph. How a real height of sinh returns exactly one real input is the next question.