Chapter 20: Inverse Trig Functions

Recall,

$arc\; cos(x)=\theta \Leftrightarrow 0\leq \theta <\pi\; and\; cos(\theta)=x$

$arc\; sin(x)=\theta \Leftrightarrow \frac{\pi}{2}\leq \theta <\frac{\pi}{2}\; and\; sin(\theta)=x$

$arc\; tan(x)=\theta \Leftrightarrow -\frac{\pi}{2}\leq \theta <\frac{\pi}{2}\; and\; tan(\theta)=x$

These are the inverse of standard trig functions.

$arc\; cot(x)=\theta \Leftrightarrow 0< \theta< \pi\; and\; cot \left ( \theta \right )=x$

$arc\; sec(x)=\theta \Leftrightarrow 0< \theta< \frac{\pi}{2}$ OR $\frac{\pi}{2}< \theta\leq \pi$ and $sec\left ( \theta \right )=x$

$arc\; cot(x)=\theta \Leftrightarrow -\frac{\pi}{2}\leq \theta <0$ OR $0< \theta\leq \frac{\pi}{2}$ and $csc\left ( \theta \right )=x$

Discuss

Where do the restrictions on $\theta$ in the definitions of $arcsec$ and $arccsc$ come from?

Answer

$sec(\theta)=\frac{1}{cos(\theta)}$ and thus $\theta \neq \frac{\pi}{2}$

$csc(\theta)=\frac{1}{sin(\theta)}$ and thus $\theta \neq 0$

What is the slope of arctan?

Example

Calculate $\frac{d}{dx}[arctan(x)]$ (The famous Formula Thirteen)

Find the derivative of $tan(x)$

$\frac{d}{dx}[tan(x)]=\frac{d}{dx}[\frac{sin(x)}{cos(x)}]$

$= \frac{cos^{2}(x)+sin^{2}(x)}{cos^{2}(x)}$

$=1/cos^{2}(x)=sec^{2}(x)$

Apply the inverse formula

$\frac{d}{dx}[arctan(x)]=\frac{1}{sec^{2}(arctan(x))}$

$=cos^{2}\left ( arctan\left ( x \right ) \right )$

Find $cos(arctan(x))$

$\theta = arctan(x)\Leftrightarrow tan\; \theta = x$

Construct a triangle with unit hypothenuse that satisfies $tan(\theta)=x$

Thus, $cos\left ( \theta \right )=\frac{1}{\sqrt{1+x^{2}}}$

$tan(\theta)=\frac{xl}{l}=x$

$l^{2}+(xl)^{2}=1$

$l^{2}(1+x^{2})=1$

^{$l=\frac{1}{\sqrt{1+x^{2}}}$}

$\frac{d}{dx}[arctan(x)]$

$=cos^{2}(arctan(x)]$

$=[\frac{1}{\sqrt{1+x^{2}}}]^{2}$

$=\frac{1}{x^{2}+1}$

Discuss

A formula for $\frac{d}{dx}[arcsin(x)]$

Find $\frac{d}{dx}[sin(x)]$

$\frac{d}{dx}[sin(x)]=cos(x))$

Apply the inverse formula

$\frac{d}{dx}[arcsin(x)]=\frac{1}{cos(arcsin(x))}$

Find $cos(arcsin(x))$

$\theta=arcsin(x)\Leftrightarrow sin(\theta)=x$

Construct a triangle with unit hypothenuse that satisfies $sin(\theta) =x$

$l^{2}+x^{2}=1^{2}$

$l=\sqrt{1-x^{2}}$

$\frac{d}{dx}[arcsin(x)]=\frac{1}{cos(arcsin(x))}=\frac{1}{l}$

$=1\sqrt{1-x^{2}}$

Discuss

Find $\frac{d}{dx}[arccos(x)]$

$\frac{d}{dx}[arccos(x)]= \frac{-1}{\sqrt{1-x^{2}}}$