Trigonometry may be the branch of mathematics that considers the relationships between lengths of sides and angles but its mathematics is almost entirely about the equality or otherwise of numerical values.

In a right angle triangle the **S**ine of an angle is the ratio between the side **O**pposite the angle and its **H**ypotenuse and the **C**osine of that angle is the ratio of the side **A**djacent to the angle and the **H**ypotenuse. Remember **SOH CAH**, your chinese friend who lives at **TOA** where **T**angents are the ratio of **O**pposites and **A**djacents.

In our diagram,

**cosθ = x/r**, **sinθ = y/r**, **tan** **θ** ** = y/x**

All are the ratios of lengths and therefore numbers so when you encounter **sin**, **cos** and **tan** in pure maths you are handling numbers not angles or lengths.

**x **and **y** are numerical lengths that lead to coordinate points on our circle. They are always at right angles and Pythagoras applies so**x ^{2} + y^{2} = r^{2}** .

It is the equation of a circle of radius r which could of course be written as

**x**

^{2}/**r**

^{2}**+ y**

^{2}/**r**

^{2}**= 1**and therefore as

**cos**

^{2}**θ + sin**

^{2}**θ = 1**

Get your calculator out and put this to the test. It is purely a numerical equality.

The following are trig equalities **sin(A ± B) = sinAcosB ± cosAsinB ****cos( A + B) = cosAcosB – sinAsinB**

**cos( A – B) = cosAcosB + sinAsinB**

Here is just one proof**The blue length value is** **sin a ****The red length value is** **cos a****Together the yellow and green length values are** **sin(a + b) **

But in the top triangle **cos b = yellow/blue**

and in the lowest triangle **sin b = green/red**

which makes **yellow = cos b sin a** *(as blue = sin a) *

and **green = sin b cos a *** ( as red = cos a)*

So **sin (a + b)** *(yellow + green )* **= cos b sin a + sin b cos a**

Again try it on your calculator. It is an identity and always true.

In any triangle like that shown the cosine rule says **c ^{2} = a^{2} + b^{2} – 2abcosC**.

Lets prove it using pythagoras

**c**

^{2}**= (b – acosC)**

^{2}**+ (asinC)**

^{2}**= b**

^{2}+ a^{2}(cos^{2}C + sin^{2}C)**– 2abcosC**

As proved above cos ** ^{2}** C + sin

**C = 1**

^{2}So

**c**The cosine rule

^{2}= a^{2}+ b^{2}– 2abcosC