# 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?

﻿$sec(\theta)=\frac{1}{cos(\theta)}$﻿ 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 ﻿$\frac{d}{dx}[arccos(x)]$﻿

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