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Level 7 - NCEA Level 2

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 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{2}$2 units. Further, because the angles in a triangle add up to $180^\circ$180° and the base angles in an isosceles triangle are equal, we can deduce that the other two unknown angles are both $45^\circ$45°.

Using our trig ratios, we can see that:

  • $\sin45^\circ=\frac{1}{\sqrt{2}}$sin45°=12
  • $\cos45^\circ=\frac{1}{\sqrt{2}}$cos45°=12
  • $\tan45^\circ=\frac{1}{1}$tan45°=11$=$=$1$1


30 and 60 degree angles

To find the exact ratios of $30$30 and $60$60 degree angles, we need to start with a 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 the length of this centre line was found using Pythagoras' theorem.

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

Using our trig ratios, we can see that:

  • $\sin30^\circ=\frac{1}{2}$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

$\sin x=\cos\left(90^\circ-x\right)$sinx=cos(90°x)


Exact value summary table

This table is another way to display the information in the exact value triangles. You can choose which method you prefer to help you remember these exact ratios.

  sin cos tan
$30^\circ$30° $\frac{1}{2}$12 $\frac{\sqrt{3}}{2}$32 $\frac{1}{\sqrt{3}}$13
$45^\circ$45° $\frac{1}{\sqrt{2}}$12 $\frac{1}{\sqrt{2}}$12 $1$1
$60^\circ$60° $\frac{\sqrt{3}}{2}$32 $\frac{1}{2}$12 $\sqrt{3}$3



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.

These particular values are ones that we need to be familiar with for our continued study in high school trigonometry, as they will help us obtain exact rather than rounded values.  Personally, I find that one of the best ways to remember them is through the use of the triangle diagrams.  

Worked Examples


Given that $\sin\theta=\frac{1}{2}$sinθ=12, we want to find the value of $\cos\theta$cosθ.

  1. First, find the value of $\theta$θ.

  2. Hence, find the exact value of $\cos30^\circ$cos30°.

Question 2

You are given that $\tan\theta=\frac{1}{\sqrt{3}}$tanθ=13.

  1. First, find the value of $\theta$θ.

  2. Hence, find exact the value of $\sin\theta$sinθ.

Question 3

Given that $\cos\theta=\frac{1}{\sqrt{2}}$cosθ=12, we want to find the value of $\tan\theta$tanθ.

  1. First find the value of $\theta$θ.

  2. Hence, find the exact value of $\tan45^\circ$tan45°.





Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions


Apply trigonometric relationships in solving problems

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