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Pythagoras’ theoremTrigonometric ratios (right-angled triangles)

Adding and subtracting fractions Simplifying surds Calculations with surds Rationalising the denominatorThis topic is relevant for:

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.

**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}} \cos(\theta)=\frac{\text{Adjacent}}{\text{Hypotenuse}} \tan(\theta)=\frac{\text{Opposite}}{\text{Adjacent}}**Step-by-step guide: **SOHCAHTOA

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}=1E.g.

Find the exact value of x.

**Method 1**

**Method 2**

Using the two right-angled triangles below, we can determine all of the exact trigonometric values we need.

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 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:

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.

In order to answer questions involving exact trig values:

**Write down the exact trig value required.****Substitute the exact trig value into the required formula/equation.****Solve the equation.****Write the answer, including the units.**

Get your free exact trig values worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free exact trig values worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEWrite down the exact value of \cos(30) .

**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 .

3**Solve the equation.**

Move on to step 4 .

4**Write 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}Write down the exact value of \sin(90)

**Write down the exact trig value required**.

We need the exact value of \sin(90)

**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 .

**Solve the equation.**

Move on to step 4 .

**Write the answer, including the units.**

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

\sin(90)=1

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

**Write down the exact trig value required**.

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

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

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

**Solve the equation.**

0+1=1 .

**Write the answer, including the units.**

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

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

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.

**Write down the exact trig value required**.

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

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

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}.

**Solve the equation.**

\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}

**Write the answer, including the units.**

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}

Calculate the length of the side x .

**Write down the exact trig value required**.

Label the triangle.

This problem involves the sine ratio:

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

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

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

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}

**Solve the equation.**

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

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

**Write the answer, including the units.**

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

Calculate the exact length of side x .

**Write down the exact trig value required**.

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:

\cos(\theta)=\frac{\text{Adjacent}}{\text{Hypotenuse}}

We need the value of \cos(45) .

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

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}

**Solve the equation.**

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}

**Write the answer, including the units.**

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

**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.

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}

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

0 \quad \quad \quad\frac{1}{2} \quad \quad \quad \frac{\sqrt{2}}{2} \quad \quad \quad 1

**(1 mark)**

Show answer

0

**(1)**

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

**(2 marks)**

Show answer

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)**

Show answer

\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)**

Show answer

Interior angle labelled 60^{\circ}

**(1)**

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

**(1)**

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

**(1)**

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

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#### FREE GCSE Maths Practice Papers - 2022 Topics

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Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

Designed to help your GCSE students revise some of the topics that will come up in the Summer exams.