Alexander Bass
This note is one of many taken during 2025

Inverse Inverse Trigonometry

In the section on derivatives of inverses, these came up

cos⁡(arcsin⁡(x))sin⁡(arccos⁡(x))sec⁡2(arctan⁡(x)) \begin{aligned} \cos(\arcsin(x))\\ \sin(\arccos(x))\\ \sec^2(\arctan(x))\\ \end{aligned}

These forms can be simplified to remove any notion of trigonometry.

Draw a right triangle with angle θ\theta, adjacent side length 1−x2\sqrt{1-x^2}, opposite side length xx, and hypotenuse side length 11.

Of course, sin⁡(θ)=x\sin(\theta) = x, so arcsin⁡(x)=θ\arcsin(x) = \theta within bounds of course.

And by definition cos⁡(θ)=1−x2\cos(\theta) = \sqrt{1-x^2}, so

cos⁡(arcsin⁡(x))=1−x2,where−1≤x≤1 \cos(\arcsin(x)) = \sqrt{1-x^2},\quad\text{where}\quad -1\leq x\leq 1

Some more identities found by similar method

cos⁡(arcsin⁡(x))=1−x2,where−1≤x≤1sin⁡(arccos⁡(x))=1−x2,where−1≤x≤1tan⁡(arcsin⁡(x))=x1−x2,where−1≤x≤1tan⁡(arccos⁡(x))=1−x2x,where−1≤x≤1 \begin{aligned} \cos(\arcsin(x)) &= \sqrt{1-x^2},\quad&\text{where}\quad &-1\leq x\leq 1\\ \sin(\arccos(x)) &= \sqrt{1-x^2},\quad&\text{where}\quad &-1\leq x\leq 1\\ \tan(\arcsin(x)) &= \frac{x}{\sqrt{1-x^2}},\quad&\text{where}\quad &-1\leq x\leq 1\\ \tan(\arccos(x)) &= \frac{\sqrt{1-x^2}}{x},\quad&\text{where}\quad &-1\leq x\leq 1\\ \end{aligned}

Arctangent is a bit more complex to deal with. Let’s do sec⁡2(arctan⁡(x))\sec^2(\arctan(x))

Draw a right triangle with angle θ\theta, adjacent side length AA, opposite side length of OO, and hypotenuse side length 11.

We know that tan⁡(θ)=OA\tan(\theta) = \frac{O}{A} so set x=OAx =\frac{O}{A}. Also known is O2+A2=1O^2+A^2 = 1

So

A2+x2A2=1A2(1+x2)=1A2=11+x2A=11+x2Ax=OO=x1+x2 \begin{aligned} A^2+x^2A^2&=1\\ A^2(1+x^2)&=1\\ A^2 &= \frac{1}{1+x^2}\\ A &= \frac{1}{\sqrt{1+x^2}}\\ Ax &= O\\ O &= \frac{x}{\sqrt{1+x^2}} \end{aligned}

And so, the triangle has been completed.

Finally, csc⁡2(θ)=1A2\csc^2(\theta) = \frac{1}{A^2}

This leads to another set of identities

sec⁡2(arctan⁡(x))=1+x2cos⁡(arctan⁡(x))=11+x2sin⁡(arctan⁡(x))=x1+x2 \begin{aligned} \sec^2(\arctan(x)) &= 1+x^2\\ \cos(\arctan(x)) &= \frac{1}{\sqrt{1+x^2}}\\ \sin(\arctan(x)) &= \frac{x}{\sqrt{1+x^2}} \end{aligned}

As a sidenote, it seems weird to me that

f(x)=sin⁡(arctan⁡(x))f−1(x)=tan⁡(arcsin⁡(x)) \begin{aligned} f(x)&= \sin(\arctan(x))\\ f^{-1}(x)&= \tan(\arcsin(x)) \end{aligned}
This note is one of many taken during 2025