## 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.

## MA101.4 Inverse Functions

Let f be a function and let $y=f(x), x \in D(f)$. We often call x the independent variable and y the dependent variable. Now consider specifying a number $y \in R(f)$ and working out what x’s are in the domain of the f, for which $f(x)=y$. There may of course be many such x’s corresponding to a particular y, but functions for which there is only ever one such x have a special name, they are called injections (their old name was one-to-one).

##### Definition

A function is an injection if for all x,y $f(x)=f(y) \Rightarrow x=y$

Or if you prefer $\forall x,y:\; x \neq y \Rightarrow f(x) \neq f(y)$

($\forall$ – for all and $\Rightarrow$ – implies.)

Injections are easily recognised from their graphs in that any line parallel to the x axis will meet the graph in at most one point.

Let f be an injection, then we use $f^{-1}(y), y \in R(y)$ to denote that single element $x \in D(f)$ which is such that $f(x)=y$.

The function $f^{-1} : R(f) \rightarrow D(f)$ is called the inverse function of the f.

By the definition of $f^{-1}$ we have, $D(f^{-1}) = R(f)$.

It is also clear that $R(f^{-1}) = D(f)$.

Note that functions that are not injections do not have inverse functions.

Example

Find the inverse function of the function f given by:

$f(x) = \begin{cases} (1 - x)^2 &\mbox{}0 \leq x \leq 1 \\ x+1 &\mbox{} 1 < x < 2 \end{cases}$

Solution: To have a sketch is a great help

$D(f)=[0,2)$ and $R(f)=[0,1] \cup (2,3)$

So $D(f^{-1}) = R(f) = [0,1] \cup (2,3)$.

Since the rule for f is given in two pieces we find the rule for $f^{-1}$ in two pieces.

$Let\; y \in [0,1] \\ So,\\ y=(1-x)^2 \\ y = 1 - 2x +x^2\\ x^2 - 2x + 1 - y = 0 \\ x = \frac{2-\sqrt{4-4(1-y)}}{2} \\ x = 1 - \sqrt{y}$

$Let\; y \in (2,3) \\So,\\ y = x+1 \; thus \; x = y-1$.

Whence $f^{-1}$ is given by

$f^{-1}(y) = \begin{cases} 1-\sqrt{y} &\mbox{}0 \leq y \leq 1 \\ y-1 &\mbox{} 2 < y < 3 \end{cases}$

Note that f is an injection and so $f^{-1}$ exists then the graphs f and $f^{-1}$ will be reflections of one another in the line y = x.