Exact Trigonometric Values

For your non-calculator paper, there are some trigonometric values you need to know. These are known as the exact trigonometric values.

The values you need to know are below, along with some helpful hints to help you remember them. You will also be shown how to work them out quickly in an exam if you are unable to remember them.

Using the Unit Triangle

This method isn’t a trick to help you remember values but it does help you to derive them. It doesn’t take very long and can be done in an exam. However, you need to understand the method for it to be useful.

To find the trigonometric values when the angle is 30° or 60°, we use an equilateral triangle that has sides of length 1cm (or 1 unit).

Exact trig values - using the unit triangle.

If we split the triangle in half, we have two right-angled triangles. For this, we will focus on the right-hand triangle. The hypotenuse of the triangle is 1cm and the base is $\frac{1}{2}$ cm (half the side of the equilateral triangle). So, by using Pythagoras’ theorem, we can find the height as $\frac{\sqrt{3}}{2}$ .

Now, by considering each angle in turn, we can find the different trig values. Consider the 30° angle; sin(30°) will be the opposite divided by the hypotenuse:

$\begin{aligned} \sin(30^{\circ}) &= \frac{1}{2} \div 1 \\ &= \frac{1}{2} \end{aligned}$

Likewise, cos(30°) will be the adjacent divided by the hypotenuse and tan(30°) will be the opposite divided by the adjacent:

$\begin{aligned} \cos(30^{\circ}) &= \frac{\sqrt{3}}{2} \div 1 \\ &= \frac{\sqrt{3}}{2} \end{aligned}$

$\begin{aligned} \tan(30^{\circ}) &= \frac{1}{2} \div \frac{\sqrt{3}}{2} \\ &= \frac{1}{\sqrt{3}} \end{aligned}$

To get the value in the table, $\frac{\sqrt{3}}{3}$ , we need to rationalise the denominator. However, it is useable in this format and $\frac{1}{\sqrt{3}}$ is accepted in exams when asked to state the exact value of tan(30°).

We can do the same when we consider the 60° angle.

$\begin{aligned} \sin(60^{\circ}) &= \frac{\sqrt{3}}{2} \div 1 \\ &= \frac{\sqrt{3}}{2} \end{aligned}$

$\begin{aligned} \cos(60^{\circ}) &= \frac{1}{2} \div 1 \\ &= \frac{1}{2} \end{aligned}$

$\begin{aligned} \tan(60^{\circ}) &= \frac{\sqrt{3}}{2} \div \frac{1}{2} \\ &= \sqrt{3} \end{aligned}$

To find the values when the angle is 45°, we consider an isosceles right-angled triangle where the two equal sides are both 1cm.

Exact trigonometric values - using Pythagoras.

Using Pythagoras, we can find the length of hypotenuse: $\sqrt{2}$ cm.

Now, we consider one of the 45° angles and find sin, cos and tan just as before:

$\begin{aligned} \sin(45^{\circ}) &= 1 \div \sqrt{2} \\ &= \frac{1}{\sqrt{2}} \end{aligned}$

$\begin{aligned} \cos(45^{\circ}) &= 1 \div \sqrt{2} \\ &= \frac{1}{\sqrt{2}} \end{aligned}$

$\begin{aligned} \tan(45^{\circ}) &= 1 \div 1 \\ &= 1 \end{aligned}$

Cover every angle by downloading yourself a copy of our Exact Trigonometric Values Walkthrough Worksheet…

The Trigonometry Hand Trick

For this trick, hold out one hand (it doesn’t matter which, but we used the left). The finger (or thumb) at the top is 0°, the next one down is 30°, then 45° and so on.

To find the trig value of a particular angle, you start by putting down the finger for that angle. So, for 30° that would be your index finger.

For sin, count how many fingers are above that finger, square root it and divide it by 2. For sin(30°) there is one finger above so:

$\begin{aligned} \sin(30^{\circ}) &= \frac{\sqrt{1}}{2} \\ &= \frac{1}{2} \end{aligned}$

For cos, count how many fingers are below that finger, square root it and divide it by 2. For cos(30°) there are three fingers below so:

$\begin{aligned} \cos(30^{\circ}) &= \frac{\sqrt{3}}{2} \end{aligned}$

For tan, square root the number fingers above and divide it by the square root of the number of fingers below. For tan(30°) there is 1 finger above and three fingers below so:

$\begin{aligned} \tan(30^{\circ}) &= \frac{\sqrt{1}}{\sqrt{3}} \\ &= \frac{1}{\sqrt{3}}\end{aligned}$