# Exact Trig Values

Here we will learn about exact trig values, including what they are, how we can derive them promptly, and how we can use them to answer questions using trigonometry.

There are also exact trig values worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are exact trig values?

Exact trig values are the exact trigonometric values for certain angles that you are expected to know for GCSE mathematics.

In trigonometry at GCSE there are three trigonometric ratios that we use, sine, cosine and tangent, though we write them as sin, cos and tan.  These trigonometric ratios show a relationship between an angle in a right-angled triangle and its side lengths.

The angle in the right-angled triangle is often labelled with a \theta (a Greek letter, ‘theta’).

\sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}

Step-by-step guide: SOHCAHTOA

### What are exact trig values? ## Exact trigonometric ratios

We use the three trigonometric ratios; sine, cosine, and tangent to calculate angles and lengths in right angled triangles. We can represent trigonometric ratios for the angles 30, 45, 60 and 90 all have exact trigonometric ratios.

We can use these exact trigonometric ratios to find lengths and angles in right angled triangles without using a calculator.

E.g.

Write down the exact value of cos 60.

cos 60 = \frac{1}{2}

E.g.

Find the exact value of sin 30 + cos 60.

sin 30 = \frac{1}{2}

cos 60 = \frac{1}{2}

sin 30 + cos 60 = \frac{1}{2}+\frac{1}{2}=1

E.g.

Find the exact value of x.

Method 1

sin 30 = \frac{5}{x}

sin 30 =\frac{1}{2}

x = 5 ÷ \frac{1}{2} = 10cm

Method 2

cos 60 = \frac{5}{x}

cos 60 =\frac{1}{2}

x = 5 ÷ \frac{1}{2} = 10cm

### Deriving the exact trig values

Using the two right-angled triangles below, we can determine all of the exact trigonometric values we need.  ### Triangle A (sometimes referred to as the 30-60 triangle)

Triangle A is half of an equilateral triangle with a side length of 2 .

We know that SOHCAHTOA is an abbreviation for the three trigonometric ratios.

Using these three ratios in the 30-60 triangle, we can determine the exact values for sin/cos/tan of 30^{\circ} , and 60^{\circ} . \text{As }\sin(\theta)=\frac{\text{Opp}}{\text{Hyp}}, \ \sin(30)=\frac{1}{2} \text{As }\sin(\theta)=\frac{\text{Opp}}{\text{Hyp}}, \ \sin(60)=\frac{\sqrt{3}}{2} \text{As }\cos(\theta)=\frac{\text{Adj}}{\text{Hyp}}, \ \cos(30)=\frac{\sqrt{3}}{2} \text{As }\cos(\theta)=\frac{\text{Adj}}{\text{Hyp}}, \ \cos(60)=\frac{1}{2} \text{As }\tan(\theta)=\frac{\text{Opp}}{\text{Adj}}, \ \tan(30)=\frac{1}{\sqrt{3}} \text{As }\tan(\theta)=\frac{\text{Opp}}{\text{Adj}}, \ \tan(60)=\frac{\sqrt{3}}{1}=\sqrt{3}

### Triangle B (sometimes referred to as the 45-45 triangle)

Triangle B is a right-angled isosceles triangle.

Using the three trig ratios in the triangle, we can determine the exact values for sin/cos/tan of 45^{\circ} . \text{As }\sin(\theta)=\frac{\text{Opp}}{\text{Hyp}}, \ \sin(45)=\frac{1}{\sqrt{2}} \text{As }\cos(\theta)=\frac{\text{Adj}}{\text{Hyp}}, \ \cos(45)=\frac{1}{\sqrt{2}} \text{As }\tan(\theta)=\frac{\text{Opp}}{\text{Adj}}, \ \tan(45)=\frac{1}{1}=1

For sin/cos/tan of 0^{\circ} and 90^{\circ} , we can use the unit circle.

Step-by-step guide: The unit circle (coming soon)

The values for sin/cos/tan of 0^{\circ} and 90^{\circ} , are as follows:

### Summary

Here is a table of the exact trig values you are expected to learn:

Note: \tan(90) is undefined.

Some of the exact values are written with a square root symbol. These numbers are called surds.  (In mathematics surds tend to be written so there are no square roots as the denominators. This is known as rationalising surds and is covered in Higher GCSE maths.) Here, we have written the non-rationalised form as these correlate with the values from the triangles above.

## How to answer questions involving exact trig values

In order to answer questions involving exact trig values:

1. Write down the exact trig value required.
2. Substitute the exact trig value into the required formula/equation.
3. Solve the equation.
4. Write the answer, including the units.

### How to answer questions involving exact trig values ## Exact trig values examples

### Example 1: stating the value

Write down the exact value of \cos(30) .

1. Write down the exact trig value required.

We need the exact value of \cos(30)

2 Substitute the exact trig value into the required formula/equation.

For this question, we do not have an equation to solve so we can move on to Step 4 .

3Solve the equation.

Move on to step 4 .

4Write the answer, including the units.

There are no defined units for this question so we can simply state the exact trig value:

\cos(30)=\frac{\sqrt{3}}{2}

### Example 2: stating the value

Write down the exact value of \sin(90)

We need the exact value of \sin(90)

For this question, we do not have an equation to solve so we can move on to Step 4 .

Move on to step 4 .

There are no defined units for this question so we can simply state the exact trig value:

\sin(90)=1

### Example 3: using the exact values

Write down the exact value of \cos(90)+\tan(45)

We need the exact value of \cos(90) and \tan(45)

As \cos(90)=0 and \tan(45)=1 , substituting these values into the equation, we have \cos(90)+\tan(45)=0+1

0+1=1 .

There are no defined units for this question so we can simply state the solution:

\cos(90)+\tan(45)=1

### Example 4: using the exact values

Write down the exact value of \sin(45)+\tan(60) . Express your answer as a single fraction with a rationalised denominator in its simplest form.

We need the exact value of \sin(45) and \tan(60)

As \sin(45)=\frac{1}{\sqrt{2}} and \tan(60)=\sqrt{3} , substituting these values into the equation, we have \sin(45)+\tan(60)=\frac{1}{\sqrt{2}}+\sqrt{3}.

\begin{aligned} \\\frac{1}{\sqrt{2}}+\sqrt{3}&=\frac{1}{\sqrt{2}}+\sqrt{3}\times\frac{\sqrt{2}}{\sqrt{2}}\\ \\&=\frac{1}{\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}}\\ \\&=\frac{1+\sqrt{6}}{\sqrt{2}}\\ \\&=\frac{\sqrt{2}(1+\sqrt{6})}{\sqrt{2}\times\sqrt{2}}\\ \\&=\frac{\sqrt{2}+\sqrt{12}}{2}\\ \\&=\frac{\sqrt{2}+2\sqrt{3})}{2}\\ \end{aligned}

There are no defined units for this question so we can simply state the solution:

\sin(45)+\tan(60)=\frac{1}{\sqrt{2}}+\sqrt{3}=\frac{\sqrt{2}+2\sqrt{3}}{2}

### Example 5: using an exact value to find a missing side

Calculate the length of the side x .

Label the triangle.

This problem involves the sine ratio:

\sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}

We need the exact value of \sin(30) .

For this question we have:

• opposite side = x
• hypotenuse = 7cm
• \sin(30)=\frac{1}{2}.

Substituting these values into \sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}} , we have

\frac{1}{2}= \frac{x}{7}

By multiplying both sides of the equation by 7 , we get

x=\frac{1}{2}\times 7=3.5 .

The length of side x=3.5\text{ cm}

### Example 6: using an exact value to find a missing side

Calculate the exact length of side x .

Labelling the triangle with the corresponding sides of the triangle in relation to the known angle, we have:

This problem therefore involves the cosine ratio:

We need the value of \cos(45) .

We now know that:

• Adjacent = 5cm
• hypotenuse = x
• \cos(45)=\frac{1}{\sqrt{2}}.

\begin{aligned} \cos(\theta)&=\frac{\text{Adjacent}}{\text{Hypotenuse}}\\\\ \frac{1}{\sqrt{2}}&=\frac{5}{x} \end{aligned}

Multiplying both sides of the equation by x , we get

\frac{x}{\sqrt{2}}=5 .

Multiplying both sides of the equation by \sqrt{2} , we get

x=5\sqrt{2}

The length of side x is 5\sqrt{2}\text{ cm}

### Common misconceptions

• Exact values

If you are asked for an exact value you must leave your answer in surd form.

There is no need to write your answer as a decimal and round.  It would no longer be an exact value.

• Using exact values

If you have been given the answer, you need to make it clear that you have used the exact trig value in your workings.

• Remembering exact sin values

The exact sin values can be difficult to remember, but there is a pattern.  Here it is.

These do simplify to the familiar, exact trig values.

• Remembering exact cos values

The exact cos values can be difficult to remember, but there is a pattern.  Here it is.

These do simplify to the familiar, exact trig values.

### Practice exact trig values questions

1. What is the exact value of \cos(60) :

\frac{\sqrt{3}}{2} \frac{1}{2} 1 \frac{\sqrt{2}}{2} \cos(60)=\frac{1}{2} 2. What is the exact value of \tan(30) :

\frac{\sqrt{3}}{3} 0 1 \sqrt{3} \tan(30)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} 3. What is the exact value of \cos(0)+\sin(30) ?

1 \frac{1+\sqrt{3}}{2} 1\frac{1}{2} \frac{1}{2} \cos(0)+sin(30)=1+\frac{1}{2}=1\frac{1}{2} 4. What is the exact value of \cos(45)+\sin(60) ?

\frac{\sqrt{3}+\sqrt{2}}{3} \frac{1+\sqrt{2}}{2} \frac{1+\sqrt{3}}{2} \frac{\sqrt{3}+\sqrt{2}}{2} \cos(45)+sin(60)=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+\sqrt{2}}{2} 5. Calculate the exact value of side x . 4.5 \text{ cm} 3 \text{ cm} 2\sqrt{3} \text{ cm} 3\sqrt{2} \text{ cm} We need to label the triangle. Then we need to be using \cos(60)=\frac{1}{2}

\begin{aligned} \cos(\theta)&= \frac{\text{Adjacent}}{\text{Hypotenuse}} \\\\ \frac{1}{2}&= \frac{x}{6} \end{aligned}

Multiplying both sides of the equation by 6 , we have

\begin{aligned} x&=6\times \frac{1}{2} \\\\ x&=3 \text{ cm} \end{aligned}

6. Calculate the exact value of side x . 4\sqrt{3} \text{ cm} \frac{\sqrt{3}}{4} \text{ cm} (4+\sqrt{3}) \text{ cm} 4.3 \text{ cm} We need to label the triangle. Then we need to be using \tan(30)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}

\begin{aligned} \tan(\theta) &= \frac{\text{Opposite}}{\text{Adjacent}} \\\\ \frac{\sqrt{3}}{3} &= \frac{x}{12} \\\\ x &= 12 \times \frac{\sqrt{3}}{3} \\\\ x &= 4 \sqrt{3} \text{ cm} \end{aligned}

### Exact trig values GCSE questions

1. Circle the value of \cos(90) .

(1 mark)

0

(1)

2. Work out the exact value of 8 \sin(45) .

(2 marks)

8\times \frac{\sqrt{2}}{2}

(1)

4\sqrt{2}

(1)

3. Use trigonometry to show that the length of side BC of this triangle is 10 \ cm . (3 marks)

\sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}

(1)

\frac{1}{2}=\frac{5}{x}

(1)

x=5\times{2}=10

(1)

4. Here is an equilateral triangle with sides 10 \ cm .

Use trigonometry to show that the perpendicular height of this triangle is 5\sqrt{3}\text{ cm} .

(3 marks) Interior angle labelled 60^{\circ}

(1)

\sin(60)=\frac{\sqrt{3}}{2}

(1)

10\times\frac{\sqrt{3}}{2}=5\sqrt{3}

(1)

## Learning checklist

You have now learned how to:

• Know the exact values of \sin \; \theta and \cos \; \theta for \theta = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ} and 90^{\circ}
• Know the exact value of \tan \; \theta for \theta = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}
• Write down the exact trig values for certain angles
• Use the exact trig values to solve problems

## The next lessons are

### Did you know?

The exact values of tan can be found by dividing the values of sin by the values of cos.

## Still stuck?

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#### GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.