Exact Trigonometric Ratios

There are some sine, cosine and tangent ratios that we are expected to know by heart for the non-calculator paper.

In this note, we will see where these exact ratios come from and use them in questions involving right-angled triangles, the sine and cosine rules, and the area of a triangle.

We will derive the exact trigonometric ratios using just two triangles.

Triangle The triangle

We begin with a right-angled triangle with a base of and an angle of

Firstly, the third angle can be found using the fact that angles in a triangle add up to

Notice that two of the angles are equal (both ), so the triangle is isosceles. This means that the height of the triangle is also

Next, we find the hypotenuse of the triangle using Pythagoras’ theorem.

The hypotenuse is represented by in the triangle below.

So,

Square rooting both sides,

Therefore, we get the following triangle.

In this triangle, the hypotenuse is the side that’s adjacent to the angle is and the side that’s opposite the angle is

With these values, we use SOHCAHTOA to find the exact values of and

To find this value, we use the SOH part of SOHCAHTOA.

Since the opposite is and the hypotenuse is

Note: can be written as if we rationalise the denominator. For more practice on this, please see our Rationalising Denominators with Surds study note.

To find this value, we use the CAH part of SOHCAHTOA.

Since the adjacent is and the hypotenuse is

Notice that this is equal to sin

To find this value, we use the TOA part of SOHCAHTOA.

Since the opposite is and the adjacent is

Therefore,

or or

Triangle The and triangle

We begin with an equilateral triangle with sides of length (you could start with any side length, however, using keeps the numbers nicer).

Firstly, we cut the triangle in half down its line of symmetry (we will call this vertical line ) to produce two congruent right-angled triangles. Every angle in an equilateral triangle is and the cut splits the top angle into two angles.

For this derivation, we only need one of the two new right-angled triangles, so we will use the triangle on the left-hand side.

Next, we find the height, using Pythagoras’ theorem.

Subtracting from both sides,

Square rooting both sides,

This triangle will give us all the and values. Let’s start with the angle.

In this triangle, the hypotenuse is the side that’s adjacent to the angle is and the side that’s opposite the angle is

With these values, we use SOHCAHTOA to find the exact values of and

To find this value, we use the SOH part of SOHCAHTOA.

Since the opposite is and the hypotenuse is

To find this value, we use the CAH part of SOHCAHTOA.

Since the adjacent is and the hypotenuse is

To find this value, we use the TOA part of SOHCAHTOA.

Since the opposite is and the adjacent is

Therefore,

Now, we move on to the angle.

This time, the hypotenuse is the side that’s adjacent to the angle is and the side that’s opposite the angle is

With these values, we use SOHCAHTOA to find the exact values of and

To find this value, we use the SOH part of SOHCAHTOA.

Since the opposite is and the hypotenuse is

To find this value, we use the CAH part of SOHCAHTOA.

Since the adjacent is and the hypotenuse is

To find this value, we use the TOA part of SOHCAHTOA.

Since the opposite is and the adjacent is

Note: Like and we can rewrite as by rationalising the denominator.

Therefore,

or

The and values

We cannot draw a right-angled triangle for these two angles. This is because a right-angled triangle already uses up so each of its other two angles must be bigger than and smaller than

Instead, we imagine a right-angled triangle with a hypotenuse of and watch what happens as the angle changes. Let’s start by decreasing the angle towards

As the angle gets closer and closer to the triangle gets flatter and flatter. Here, the opposite side shrinks towards while the adjacent side grows towards the hypotenuse, which is

So, using SOHCAHTOA, with as the opposite, as the adjacent, and as the hypotenuse:

Therefore,

Now, we move onto the values. Consider the right-angled triangle with a hypotenuse of and, this time, watch what happens as the angle increases towards

As the angle gets closer and closer to the triangle gets taller and thinner. This time, the opposite side grows towards the hypotenuse, and the adjacent side shrinks towards

So, using SOHCAHTOA, with as the opposite, as the adjacent, and as the hypotenuse:

For tan we would need

However, dividing by is not possible, so tan is undefined.

Note: We can see this on a calculator. For instance, and As the angle gets closer to just keeps growing, so there is no value we can give to If we type in the calculator, we’d get an error message.

Therefore,

undefined

Below is a table which summarises all the exact trigonometric values that we need to know by heart.

The Exact Trigonometric Ratios

Here is how we could remember the values in this table:

For the row, we can use the pattern shown below: the number inside the square root increases from to and each value is divided by

The row is just the sin row written backwards.

Each value can be found by doing for that angle.

This reinforces why is undefined, because we would be doing

which, as mentioned before, is not possible.

In this section, we will practise using our exact ratios when working with right-angled triangles. These types of questions appear often in non-calculator papers.

The exact trigonometric ratios also appear in non-calculator questions about triangles that are not right- angled. The rules and methods stay the same, but we simply substitute the exact values instead of using a calculator.

Here is a reminder of the rules that we’ve learnt. For more information on them, please read our Cosine Rule, Sine Rule, and Areas of Non-Right-Angled Triangles study notes.

The cosine rule is given by

The sine rule is given by

The area, , of a triangle is given by

In this section, we will practise harder exam-style questions which combine exact ratios and algebra. For these questions, we’d form equations, using the rules and exact ratios, and then solve them.