Figure 7.1 shows a triangle with sides a, b and c and angles subtending the sides A, B and C respectively.

Figure 7.1: Triangle with side lengths a, b and c.

For the triangle in Figure 7.1 we can write

$\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$

These equations are called the sine rule.

For the same triangle we can also write

$a^2=b^2+c^2-2bc\ cos A$

This equation is called the cosine rule. When angle $A=\pi/2$, so we have a right angled triangle and the cosine rule simplifies to Pythagoras's theorem.

$a^2=b^2+c^2$

For right angled triangles we can write $cos \theta= \frac{x}{r}$ and $sin \theta= \frac{y}{r}$. Imagine standing at the angle $ \theta $ and look along the x-axis which is the side of the triangle. We could say we were looking in the central direction and the central direction gives us $ cos \theta = x/r$.

The line that is perpendicular to the x-axis goes sideways and sideways gives us $ sin \theta = y/r$.

Exercise 7.1: For the triangle in figure 7.1 calculate the following:

a) a = 4, b = 5 and c = 6. Find all the angles.

b) a = 5, b = 9 and c = 7. Find all the angles.

c) a = 3, b = 5 and C = 40°. Find the length of side c.

d) b = 2, c = 6 and A = 58°. Find the length of side a.

So far we have been sketching our diagrams on x-y axes. This system is attributed to René Descartes and is called the **Cartesian coordinate system**. The system allows us to describe any point on a 2D plane.

An alternative way to describe a point on a 2D plane is by giving an angle from a fixed line and a radius. This system is calle the **polar coordinate system** and is used a lot in navigation.

To convert from cartesian coordinates to polar coordinates we have:

$r^2=x^2+y^2$

$\theta = tan^{-1}\ y/x$

To convert from polar coordinates to cartesian coordinates we have:

$x=r\ cos\ \theta$

$y=r\ sin\ \theta$

Exercise 7.2: For the triangle in figure 7.1 calculate the following:

a) a = 4, b = 5 and c = 6. Find all the angles.

b) a = 5, b = 9 and c = 7. Find all the angles.

c) a = 3, b = 5 and C = 40°. Find the length of side c.

d) b = 2, c = 6 and A = 58°. Find the length of side a.

Imagine a disc with a pen mounted near the edge. If the disc rotates about its centre the pen will draw a circle. If we now pull the paper from under the disc the pen will draw a sinusoidal curve and the type of curve will depend on the direction we pull the paper.

Pythagoras's theorem states $x^2+y^2=r^2$. If we keep $r$ constant and allow $x$ to vary from $x=r$ through $x=0$ to $x=-r$ we get

If we divide both sides by $r^2$ we get

$(\frac{x}{r})^2+(\frac{y}{r})^2=1$.

Looking at figure 7.1 we can see $cos \theta = \frac{x}{r}$ and $sin \theta = \frac{y}{r}$ so we can change our equation to:

$cos^2 \theta+sin^2 \theta=1$.

Here are some other trigonometric identities

$ \cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi $

$ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi $

$ \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi $

$ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi $

$ \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1-\tan\theta \tan\phi} $

$ \tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1+\tan\theta \tan\phi} $

If we let $\theta = \phi$ we get

$ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta $

$ \sin(2\theta) = 2 \sin \theta \cos \theta $

$ \tan(2\theta) = \frac{2\tan \theta}{1-\tan^2 \theta} $