You'll have noticed by now that when you find angles using trigonometric ratios, you often get long decimal answers. If, for example, you put $\cos30^\circ$cos30° into the calculator, you will see an answer of $0.86602$0.86602... which we'd have to round. However, when you take cos, sin or tan of some angles, you can express the answer as an exact number, rather than a decimal. It just may include irrational numbers. We often use these exact ratios in relation to $30^\circ$30°, $45^\circ$45° and $60^\circ$60°.

Let's look at how to do this now.

Exact value triangles

$45$45 degree angles

Below is a right-angle isosceles triangle, with the equal sides of $1$1 unit. Using Pythagoras' theorem, we can work out that the hypotenuse is $\sqrt{1^2+1^2}=\sqrt{2}$√12+12=√2 units. The angles in a triangle add up to $180^\circ$180° and the base angles in an isosceles triangle are equal, so we can also work out that the two unknown angles are both $45^\circ$45°.

We can then use our trig ratios to determine the exact values of the following:

$\sin45^\circ=\frac{1}{\sqrt{2}}$sin45°=1√2
$\cos45^\circ=\frac{1}{\sqrt{2}}$cos45°=1√2
$\tan45^\circ=\frac{1}{1}$tan45°=11$=$=$1$1

$30$30 and $60$60 degree angles

To find the exact ratios of $30$30 and $60$60 degree angles, we need to start with an equilateral triangle with side lengths of $2$2 units. Remember all the angles in an equilateral triangle are $60^\circ$60°.

Then we are going to draw a line that cuts the triangle in half into two congruent triangles. The base line is cut into two $1$1 unit pieces and cuts one of the $60^\circ$60° angles in half.

Now let's just focus on one half of this triangle.

We can calculate the length of the centre line to be $\sqrt{3}$√3 using Pythagoras' theorem. We can then use our trig ratios to determine the exact values of the following:

$\sin30^\circ=\frac{1}{2}$sin30°=12
$\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32

$\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32
$\cos60^\circ=\frac{1}{2}$cos60°=12

Notice that the $\sin60^\circ=\cos30^\circ$sin60°=cos30°. It is true for any two complementary angles that $\sin x=\cos\left(90^\circ-x\right)$sinx=cos(90°−x).

Observations

Now, an isosceles right-angled triangle may not have its sides measuring $1$1,$1$1 and $\sqrt{2}$√2, but however large it is, it will always have two $45^\circ$45° angles and the ratios of the sides will always be the same as in the table. The same applies to the triangle with $60^\circ$60° and $30^\circ$30° angles. As long as a triangle is similar to one of these triangles (it has the same angles) we can use the exact values.

Exact values

We have found the exact values for the following using the $45$45 and $30$30 $60$60 triangles:

	sin	cos	tan
$30^\circ$30°	$\frac{1}{2}$12	$\frac{\sqrt{3}}{2}$√32	$\frac{1}{\sqrt{3}}$1√3
$45^\circ$45°	$\frac{1}{\sqrt{2}}$1√2	$\frac{1}{\sqrt{2}}$1√2	$1$1
$60^\circ$60°	$\frac{\sqrt{3}}{2}$√32	$\frac{1}{2}$12	$\sqrt{3}$√3

We also have the exact values that don't describe physical triangles, at $0^\circ$0° and $90^\circ$90°:

	sin	cos	tan
$0^\circ$0°	$0$0	$1$1	$0$0
$90^\circ$90°	$1$1	$0$0	undefined

Practice questions

QUESTION 1

Use the exact value triangles in the diagram below to answer the following:

What is the exact value of $\cos45^\circ$cos45°?
What is the exact value of $\cos60^\circ$cos60°?
What is the exact value of $\cos30^\circ$cos30°?

Question 2

$\theta$θ is an angle in a right-angled triangle where $\tan\theta=\frac{1}{\sqrt{3}}$tanθ=1√3.

Solve for the exact value of $\theta$θ.

20.10 Exact values (Extended)