7.01 Exact values
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.
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.