## MA101.13 Trigonometrical functions

Series definitions for the sin and cosine functions are:

$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$
$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...$

These converge $\forall x \in \mathbb{R}$.

If we differentiate these term by term we can see that:

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

Many other properties can be deduced from these power series.

The graph of $y = \sin{x}$ is as shown:

We can see that sin is not an injection (domain of $\mathbb{R}$) and so there is no inverse. However the function $f:[-\frac{\pi}{2},\frac{\pi}{2}]\to[-1,1]$, or $f:x\to \sin{x}$ [called the cut down sine], has the graph:

and this is an injection, and has an inverse function $f^{-1}$ with domain $latex[-1,1]$ and range $[-\frac{\pi}{2},\frac{\pi}{2}]$. $f^{-1}$ is the unique real number (angle) between $[-\frac{\pi}{2},\frac{\pi}{2}]$ whose sine is x. The $f^{-1}$ is symbolised by $\sin^{-1}{x}$ or $\arcsin{x}$.

The graph of $\arcsin{x}$ is a reflection of $y=\sin{x}$ (cut down) in the line $y=x$.

Knowing $y=\arcsin{x}$ is true, you may deduce $x=\sin{x}$ is true.

Example: $\frac{\pi}{4}=arcsin(\frac{1}{\sqrt{2}}) \Rightarrow sin)\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Knowing $x = sin(y)$ is true, you may not deduce $y = arcsin(x)$.

Example: $\frac{1}{\sqrt{2}}=sin(\frac{3\pi}{4}) \nRightarrow \frac{3\pi}{4} = arcsin(\frac{1}{\sqrt{2}})$.

##### Theorem

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

###### Proof

\begin{aligned} \mbox{Let: } y &= arcsin(x) \\ \mbox{then } x &= sin(y) \\ \frac{dx}{dy} &= cos(y) \\ \frac{dy}{dx} &= \frac{1}{cos(y)} \\ &= \frac{1}{\sqrt{1-sin^2(y)}} \\ &= \frac{1}{\sqrt{1-x^2}} \end{aligned}

We take the positive square root because for $[-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}]$ the $\cos{y} \geq 0$

Similarly we define the ‘so called’ cut down cosine – this is cosine but with the domain $[0,\pi]$, and the cut down tangent with domain $(-\frac{\pi}{2}, \frac{\pi}{2})$. The inverses of these functions are arccos and arctan.

## MA101.12 The exponential functions

We define

$\exp{x} = 1 +x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...$ which in fact converges $\forall x \in \mathbb{R}$.

\begin{aligned} \exp{x} \cdot \exp{y} &= (1 +x+\frac{x^2}{2!}+...)(1 +y+\frac{y^2}{2!}+...) \\&= 1 +x+y+\frac{x^2}{2!}+xy+\frac{y^2}{2!}+...+\frac{x^n}{n!}+\frac{x^{n-1}y}{n-1!}+\frac{x^{n-2}y^2}{(n-2)!2!}+...+\frac{y^2}{2!}+... \\&= 1 +(x+y)+\frac{(x+y)^2}{2!}+...+\frac{(x+y)^n}{n!}+... \\&= \exp{(x+y)} \end{aligned}

We have assumed that we can multiply infinite series as if they were finite algebraic expressions. This is not always true, but it is for convergent power series.

The relationship $\exp{x}\cdot\exp{y} = \exp{(x+y)}$ leads us to write $\exp{x}=e^x$ so that $e^x\cdot e^y = e^{x+y}$.

Note, $\exp{1} = e^1 = e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...$

On the assumption that a power series can be differentiated term by term $\frac{d e^x}{dx} = 0 + 1 +\frac{x^2}{2!}+...=e^x$.

##### Properties of exp

The graph of $y=e^x$ is as shown, (with series definition this is not easy to see), but there is some evidence from the property $e^{x_1}\cdot e^{x_2}=e^{x_1+x_2}$

Observe from the graph the properties:

1. $\displaystyle \lim_{x\to\infty}e^x=\infty$.
2. $\displaystyle \lim_{x\to-\infty}e^x=0$.
3. $\forall x, e^x >0$.
4. $e^x$ is continuous everywhere.
5. $D(e^x)=\mathbb{R}$.
6. $e^x$ is an injection (or one to one).
ie. $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

This means there will be an inverse function. We call it the log function. We have $\log{y}$ equal to the unique x such that $y=e^x$.

$y=e^x$ and $x=\ln{y}$ mean of course the same thing. The graph of $y=\ln{x}$ will be a reflection of $y=e^x$ in the line $y=x$.

Note that the domain of ln = range of exp = $\mathbb{R}^+$.

##### Properties of ln
1. If $y=\ln{x}$ then $x=e^y$, so $\frac{dx}{dy} = e^y$.
Therefore, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^y} = \frac{1}{x}$.
2. $\ln{x}$ is an injection.
Proof:
Let $\ln{x}=\ln{y}$.
Put $\ln{x}=p$ and $\ln{y}=p$.
So, $e^p=x$ and $e^p=y$.
Whence $x=y$.
3. $\forall x, x=e^{\ln{x}}$.
Proof:
Let $y=e^{ln{x}}$.
Then $\ln{y} = \ln{x}$.
Whence $x=y$ by 2.
So, $x=e^{\ln{x}}$.
4. $\ln{xy}=\ln{x}+\ln{y}$.
Proof:
\begin{aligned} \ln{xy}&=\ln{e^{\ln{x}}\cdot e^{\ln{y}}} \\&= \ln{e^{\ln{x}+\ln{y}}} \\ xy &= e^{ln{x}+\ln{y}} \\ \mbox{Hence}\; \ln{x}+\ln{y} &= \ln{xy} \end{aligned}.
5. $\forall a \in \mathbb{R}^+$
(i) if  m in an integer then $a^m = m\ln{a}$.
(ii) if $\frac{m}{n}$ is in $\mathbb{Q}$, then equally we have $\ln{a\frac{m}{n}}=\frac{m}{n}\ln{a}$. ($a^{\frac{m}{n}} \mbox{ means } \sqrt[n]{a^m}$.
##### Definition

For $a \in \mathbb{R}^+$ and $x \in \mathbb{R}$ we define $a^x = e^{x\ln{a}}$.

For example $3^{\sqrt{2}}=e^{\sqrt{2}\ln{3}}$.

Note that this definition fits in with the usual one when x is rational.

For fixed $a \in \mathbb{R}^+, a^x$ is a function.

Note: $\frac{d(a^x)}{dx} = \frac{d(e^{\ln{a^x}}}{dx} = \frac{d(e^{x\ln{a}})}{dx} = \ln{a(e^{x\ln{a}})} = a^x\ln{a}$.