MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 30 July 2014

Lecture 7

POOF! ... and there etched in stone was $$e^{ix}=\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x\text{.}$$ Do you believe it?

Verify $$\begin{array}{ccc}{\displaystyle \text{cos}(x+y)}& =& {\displaystyle \text{cos}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y-\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}y,}\\ {\displaystyle \text{sin}(x+y)}& =& {\displaystyle \text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y+\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y\text{.}}\end{array}$$ Well $$\begin{array}{ccc}{\displaystyle e^{ix}{e}^{iy}}& =& {\displaystyle (\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x)(\text{cos}\hspace{0.17em}y+i\text{sin}\hspace{0.17em}y)}\\ & =& {\displaystyle \text{cos}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y+i\text{cos}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}y+i\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y+i^2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}y}\\ & =& {\displaystyle \text{cos}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y-\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}y+i(\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y+\text{cos}\hspace{0.17em}x+\text{sin}\hspace{0.17em}y)\text{.}}\end{array}$$ On the other hand $$e^{ix}{e}^{iy}=e^{i(x+y)}=\text{cos}(x+y)+i\text{sin}(x+y)\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \text{cos}(x+y)}& =& {\displaystyle \text{cos}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y-\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}y\text{and}}\\ {\displaystyle \text{sin}(x+y)}& =& {\displaystyle \text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y+\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}y\text{.}}\end{array}$$

Verify $$\begin{array}{ccc}{\displaystyle \text{sin}\hspace{0.17em}x}& =& {\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots ,}\\ {\displaystyle \text{cos}\hspace{0.17em}x}& =& {\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \text{.}}\end{array}$$ Well $$\begin{array}{ccc}{\displaystyle e^{ix}}& =& {\displaystyle 1+ix+\frac{{\left(ix\right)}^2}{2!}+\frac{{\left(ix\right)}^3}{3!}+\frac{{\left(ix\right)}^4}{4!}+\frac{{\left(ix\right)}^5}{5!}+\cdots }\\ & =& {\displaystyle 1+ix+\frac{i^2{x}^2}{2!}+\frac{i^3{x}^3}{3!}+\frac{i^4{x}^4}{4!}+\frac{i^5{x}^5}{5!}+\frac{i^6{x}^6}{6!}+\cdots }\\ & =& {\displaystyle \begin{array}{c}1\phantom{+ix}-\frac{x^2}{2!}\phantom{-\frac{i{x}^3}{3!}}+\frac{x^4}{4!}\phantom{+\frac{i{x}^5}{5!}}-\frac{x^6}{6!}\phantom{-\frac{i{x}^7}{7!}}+\cdots \\ \phantom{1}+ix\phantom{-\frac{x^2}{2!}}-\frac{i{x}^3}{3!}\phantom{+\frac{x^4}{4!}}+\frac{i{x}^5}{5!}\phantom{-\frac{x^6}{6!}}-\frac{i{x}^7}{7!}+\cdots \end{array}}\\ & =& {\displaystyle (1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots )+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )\text{.}}\end{array}$$ On the other hand $$e^{ix}=\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \text{cos}\hspace{0.17em}x}& =& {\displaystyle (1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots )\text{and}}\\ {\displaystyle \text{sin}\hspace{0.17em}x}& =& {\displaystyle (x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )\text{.}}\end{array}$$

Verify $\text{sin}(-x)=-\text{sin}\hspace{0.17em}x$ and $\text{cos}(-x)=\text{cos}\hspace{0.17em}x\text{.}$ $$\begin{array}{ccc}{\displaystyle \text{sin}(-x)}& =& {\displaystyle (-x)-\frac{{(-x)}^3}{3!}+\frac{{(-x)}^5}{5!}-\frac{{(-x)}^7}{7!}+\cdots }\\ & =& {\displaystyle -x-\frac{-x^3}{3!}+\frac{-x^5}{5!}-\frac{-x^7}{7!}+\cdots }\\ & =& {\displaystyle -x-\frac{-x^3}{3!}+\frac{-x^5}{5!}-\frac{-x^7}{7!}+\cdots }\\ & =& {\displaystyle -(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )=-\text{sin}\hspace{0.17em}x\text{.}}\\ {\displaystyle \text{cos}(-x)}& =& {\displaystyle 1-\frac{{(-x)}^2}{2!}+\frac{{(-x)}^4}{4!}-\frac{{(-x)}^6}{6!}+\frac{{(-x)}^8}{8!}-\cdots }\\ & =& {\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots }\\ & =& {\displaystyle \text{cos}\hspace{0.17em}x\text{.}}\end{array}$$

Verify ${\text{sin}}^{2}x+{\text{cos}}^{2}x=1\text{.}$

Well $$e^{ix}{e}^{-ix}=e^{ix-ix}=e^0=1\text{.}$$ On the other hand $$\begin{array}{ccc}{\displaystyle e^{ix}{e}^{-ix}}& =& {\displaystyle (\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x)(\text{cos}(-x)+i\text{sin}(-x))}\\ & =& {\displaystyle (\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x)(\text{cos}\hspace{0.17em}x-i\text{sin}\hspace{0.17em}x)}\\ & =& {\displaystyle {\text{cos}}^{2}x-i\text{cos}\hspace{0.17em}x\hspace{0.17em}\text{sin}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x-i^2{\text{sin}}^{2}x}\\ & =& {\displaystyle {\text{cos}}^{2}x+{\text{sin}}^{2}x\text{.}}\end{array}$$ So $$\begin{array}{|c|}\hline 1={\text{cos}}^{2}x+{\text{sin}}^{2}x\\ \hline\end{array}$$

Find $\frac{dy}{dx}$ when $y=\text{sin}\hspace{0.17em}x\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{d\text{sin}\hspace{0.17em}x}{dx}}\\ & =& {\displaystyle \frac{d(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )}{dx}}\\ & =& {\displaystyle 1-\frac{3x^2}{3·2·1}+\frac{5x^4}{5·4·3·2}-\frac{7x^6}{7·6·5·4·3·2·1}+\cdots }\\ & =& {\displaystyle 1-\frac{x^2}{2·1}+\frac{x^4}{4·3·2}-\frac{x^6}{6·5·4·3·2}+\cdots }\\ & =& {\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots }\\ & =& {\displaystyle \text{cos}\hspace{0.17em}x\text{.}}\end{array}$$ So $$\frac{d\text{sin}\hspace{0.17em}x}{dx}=\text{cos}\hspace{0.17em}x\text{.}$$

Find $\frac{d\text{cos}\hspace{0.17em}x}{dx}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{d\text{cos}\hspace{0.17em}x}{dx}}& =& {\displaystyle \frac{d(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+\cdots )}{dx}}\\ & =& {\displaystyle -\frac{2x}{2·1}+\frac{4x^3}{4·3·2·1}-\frac{6x^5}{6·5·4·3·2}+\frac{8x^7}{8·7·6·5·4·3·2}-\cdots }\\ & =& {\displaystyle -x+\frac{x^3}{3·2·1}-\frac{x^5}{5·4·3·2}+\frac{x^7}{7·6·5·4·3·2}-\cdots }\\ & =& {\displaystyle -x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\cdots }\\ & =& {\displaystyle -(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )}\\ & =& {\displaystyle -\text{sin}\hspace{0.17em}x\text{.}}\end{array}$$ So $$\frac{d\text{cos}\hspace{0.17em}x}{dx}=-\text{sin}\hspace{0.17em}x\text{.}$$

Inverse trig functions

The inverse function to $f$ is the function that undoes $f\text{.}$ ${\text{sin}}^{-1}x$ is the inverse function to $\text{sin}\hspace{0.17em}x\text{.}$
${\text{cos}}^{-1}x$ is the inverse function to $\text{cos}\hspace{0.17em}x\text{.}$
${\text{tan}}^{-1}x$ is the inverse function to $\text{tan}\hspace{0.17em}x\text{.}$
${\text{cot}}^{-1}x$ is the inverse function to $\text{cot}\hspace{0.17em}x\text{.}$
${\text{sec}}^{-1}x$ is the inverse function to $\text{sec}\hspace{0.17em}x\text{.}$
${\text{csc}}^{-1}x$ is the inverse function to $\text{csc}\hspace{0.17em}x\text{.}$ So $${\text{sin}}^{-1}\left(\text{sin}\hspace{0.17em}B\right)=B\text{and}\text{sin}\left({\text{sin}}^{-1}x\right)=x\text{.}$$ If $y=\text{sin}\hspace{0.17em}x$ then ${\text{sin}}^{-1}y=x\text{.}$ Since $0=\text{sin}\hspace{0.17em}0$ then ${\text{sin}}^{-1}0=0,$ and $1=\text{sin}\hspace{0.17em}\frac{\pi }{2}$ so ${\text{sin}}^{-1}\hspace{0.17em}1=\frac{\pi }{2}\text{.}$

Warnings

(1)

$f\left(x\right)={\text{sin}}^{-1}x$ is not a function in the strictest sense. $${\text{sin}}^{-1}0=0\text{or}\pi \text{or}2\pi \text{or}-\pi \text{or}-2\pi \text{or}10\pi \text{or}\dots$$ This is similar to $\sqrt{9},$ which is $3$ or $-3$ depending on the context.

(2)

$${\text{cos}}^{-1}x\ne {\left(\text{cos}\hspace{0.17em}x\right)}^{-1}\text{.}$$ ${\text{cos}}^{-1}x$ is the function that undoes $\text{cos}\hspace{0.17em}x\text{.}$ $${\left(\text{cos}\hspace{0.17em}x\right)}^{-1}=\frac{1}{\text{cos}\hspace{0.17em}x}\text{.}$$ For example: $$\begin{array}{ccc}{\displaystyle {\left(\text{cos}\hspace{0.17em}0\right)}^{-1}}& =& {\displaystyle \frac{1}{\text{cos}\hspace{0.17em}0}=\frac{1}{1}=1,\text{and}}\\ {\displaystyle {\text{cos}}^{-1}0}& =& {\displaystyle \frac{\pi }{2}\text{or}\frac{-\pi }{2}\text{or}\frac{3\pi }{2}\text{or}\frac{-3\pi }{2}\text{or}\cdots }\end{array}$$ but ${\text{cos}}^{-1}0$ is never equal to $1\text{.}$

Verify $\text{sin}\left({\text{cot}}^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}\text{.}$

Let $y={\text{cot}}^{-1}x\text{.}$ Then $\text{cot}\hspace{0.17em}y=x\text{.}$ $$\begin{array}{c} 1+x2 x y 1 \end{array}$$ So $$\text{sin}\left({\text{cot}}^{-1}x\right)=\text{sin}\hspace{0.17em}y=\frac{1}{\sqrt{1+x^2}}\text{.}$$

Verify ${\text{sin}}^{-1}(-x)=-{\text{sin}}^{-1}x\text{.}$ $${\displaystyle {\text{sin}}^{-1}(-x)}& \stackrel{?}{=}& {\displaystyle -{\text{sin}}^{-1}x}\\ {\displaystyle \text{sin}\left({\text{sin}}^{-1}(-x)\right)}& \stackrel{?}{=}& {\displaystyle \text{sin}(-{\text{sin}}^{-1}x)}\\ {\displaystyle -x}& \stackrel{?}{=}& {\displaystyle -\text{sin}\left({\text{sin}}^{-1}x\right)}\\ {\displaystyle -x}& \stackrel{?}{=}& {\displaystyle -x}$$ YES!!

Verify ${\text{tan}}^{-1}x={\text{cot}}^{-1}\left(\frac{1}{x}\right)\text{.}$

Say $y={\text{tan}}^{-1}x\text{.}$ Then $\text{tan}\hspace{0.17em}y=x\text{.}$ So $${\text{cot}}^{-1}\left(\frac{1}{x}\right)={\text{cot}}^{-1}\left(\frac{1}{\text{tan}\hspace{0.17em}y}\right)={\text{cot}}^{-1}\left(\text{cot}\hspace{0.17em}y\right)=y={\text{tan}}^{-1}x\text{.}$$

Find $\frac{d\text{tan}\hspace{0.17em}x}{dx}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{d\text{tan}\hspace{0.17em}x}{dx}}& =& {\displaystyle \frac{d\left(\frac{\text{sin}\hspace{0.17em}x}{\text{cos}\hspace{0.17em}x}\right)}{dx}}\\ & =& {\displaystyle \frac{d\left(\text{sin}\hspace{0.17em}x\right){\left(\text{cos}\hspace{0.17em}x\right)}^{-1}}{dx}}\\ & =& {\displaystyle \text{sin}\hspace{0.17em}x\frac{d{\left(\text{cos}\hspace{0.17em}x\right)}^{-1}}{dx}+\frac{d\text{sin}\hspace{0.17em}x}{dx}{\left(\text{cos}\hspace{0.17em}x\right)}^{-1}}\\ & =& {\displaystyle \text{sin}\hspace{0.17em}x(-1){\left(\text{cos}\hspace{0.17em}x\right)}^{-2}\frac{d\text{cos}\hspace{0.17em}x}{dx}+\text{cos}\hspace{0.17em}x\frac{1}{\text{cos}\hspace{0.17em}x}}\\ & =& {\displaystyle \frac{(-\text{sin}\hspace{0.17em}x)}{{\text{cos}}^{2}x}(-\text{sin}\hspace{0.17em}x)+1=\frac{{\text{sin}}^{2}x+{\text{cos}}^{2}x}{{\text{cos}}^{2}x}}\\ & =& {\displaystyle \frac{1}{{\text{cos}}^{2}x}}\\ & =& {\displaystyle {\text{sec}}^{2}x\text{.}}\end{array}$$

Find $\frac{d{\text{tan}}^{-1}x}{dx}\text{.}$

Set $y={\text{tan}}^{-1}x\text{.}$ Then $\text{tan}\hspace{0.17em}y=x\text{.}$ Take the derivative: $${\text{sec}}^{2}y\frac{dy}{dx}=1\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{1}{{\text{sec}}^{2}y}={\text{cos}}^{2}y}\\ & =& {\displaystyle {\left(\frac{1}{\sqrt{1+x^2}}\right)}^2=\frac{1}{1+x^2}\text{.}}\end{array}\begin{array}{c} 1+x2 1 y x \end{array}$$

Find $\frac{d\text{sec}\hspace{0.17em}x}{dx}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{d\text{sec}\hspace{0.17em}x}{dx}}& =& {\displaystyle \frac{d\frac{1}{\text{cos}\hspace{0.17em}x}}{dx}=\frac{d{\left(\text{cos}\hspace{0.17em}x\right)}^{-1}}{dx}=-{\left(\text{cos}\hspace{0.17em}x\right)}^{-2}\frac{d\text{cos}\hspace{0.17em}x}{dx}}\\ & =& {\displaystyle -\frac{1}{{\text{cos}}^{2}x}(-\text{sin}\hspace{0.17em}x)=\frac{\text{sin}\hspace{0.17em}x}{{\text{cos}}^{2}x}=\frac{\text{sin}\hspace{0.17em}x}{\text{cos}\hspace{0.17em}x}·\frac{1}{\text{cos}\hspace{0.17em}x}}\\ & =& {\displaystyle \text{tan}\hspace{0.17em}x\hspace{0.17em}\text{sec}\hspace{0.17em}x\text{.}}\end{array}$$

Find $\frac{d{\text{sec}}^{-1}x}{dx}\text{.}$

Let $y={\text{sec}}^{-1}x\text{.}$ Then $\text{sec}\hspace{0.17em}y=x\text{.}$ So $$\text{tan}\hspace{0.17em}y\hspace{0.17em}\text{sec}\hspace{0.17em}y\frac{dy}{dx}=1\text{.}$$ So $$\frac{dx}{dx}=\frac{1}{\text{tan}\hspace{0.17em}y\hspace{0.17em}\text{sec}\hspace{0.17em}y}\text{.}$$ $$\begin{array}{c} x 1 y 1+x2 \end{array}$$ So $$\frac{dy}{dx}=\frac{1}{\sqrt{x^2-1}x}\text{.}$$ So $$\frac{d{\text{sec}}^{-1}x}{dx}=\frac{1}{x\sqrt{x^2-1}}\text{.}$$

Notes and References

These are a typed copy of Lecture 7 from a series of handwritten lecture notes for the class MATH 221 given on September 20, 2000.

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