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}}).


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


\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).


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.


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.