Exact Trig Values (Cambridge (CIE) IGCSE International Maths)

Revision Note

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Expertise

Maths

For certain angles the values of sin θ, cos θ and tan θ can be written exactly
- This means using fractions and surds
You are expected to know the exact values of sin, cos and tan for
- 0°, 30°, 45°, 60°, 90°, 180° and their multiples

Look at patterns in the table
- Note the values of sin θ from 0° to 90° match cos θ in reverse, from 90° to 0°
- Some people remember sin θ using the trick $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ which simplifies to $0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1$
Two special right-angled triangles below can help you to find the exact values for 30º, 45º and 60º
- Remember that by rationalising the denominator, $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

The trig graphs can help you to remember the exact values for 0º and multiples of 90º

You may come across trig questions in a non-calculator question
In trig calculations, substitute in the exact trig values and solve as usual
- E.g. Solve the equation $\cos 45 = \frac{x}{12}$
  - Replace $\cos 45$ with $\frac{\sqrt{2}}{2}$ to give $\frac{\sqrt{2}}{2} = \frac{x}{12}$
  - Then you can solve for $x$
On trig graphs, you may be expected to find a coordinate
- E.g. The coordinates $(30, k)$ lie on the graph $y = \tan x$ , find $k$
  - $k$ will be equal to $\tan 30$
  - The exact value of $\tan 30$ is $\frac{\sqrt{3}}{3}$
  - Therefore $k = \frac{\sqrt{3}}{3}$

Writing these out (or sketching the triangles/graphs) on your paper at the beginning of the exam means that you can use them as many times as you need to during the exam!

Find the value of $x$ in the diagram below.

Give your answer as an exact value.

Triangle ABC is a right-angled triangle, so use SOHCAHTOA

We know the hypotenuse (AC) and we want to calculate the opposite (BC), so use $\sin θ = \frac{O}{H}$

$\sin 60 = \frac{x}{28}$

Remember that $\sin 60 = \frac{\sqrt{3}}{2}$

So,

$\begin{array}{rcl} \frac{\sqrt{3}}{2} & = & \frac{x}{28} \\ 28 \times \frac{\sqrt{3}}{2} & = & x \\ 14 \times \sqrt{3} & = & x \end{array}$

Leave in exact (surd) form

$x = 14 \sqrt{3}$ cm

the (exam) results speak for themselves:

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