# Trigonometry Common Values Revision and Worksheets

GCSE 6 - 7AQAEdexcelOCRWJECHigherAQA 2022Edexcel 2022OCR 2022WJEC 2022

## Trigonometry Common Values

Up to this point, we’ve let our calculators do the heavy lifting when it comes to evaluating the $3$ trigonometry functions, but there are some values of the trigonometry functions that you have to remember.

Make sure you are happy with the following topics before continuing:

Level 6-7 GCSE    ## Values You need to learn

Specifically, you will have to remember:

$\sin$, $\cos$, and $\tan$ of $0\degree, 30\degree, 45\degree, 60\degree,$ and $90\degree$ You need to remember all of these values.

Level 4-5 GCSE    ## Example

Find the value of $y$ shown on the right-angled triangle.

[2 marks] First we have to find which equation we need to use.

The hypotenuse $=\textcolor{blue}{10}$ cm

The adjacent $= \textcolor{limegreen}{y}$

This means we use $\textcolor{purple}{C}\textcolor{limegreen}{A}\textcolor{blue}{H}$

$\textcolor{purple}{\cos(30\degree)} = \dfrac{\textcolor{limegreen}{y}}{\textcolor{blue}{10}}$

We can see in the table that $\textcolor{purple}{\cos(30\degree)} = \textcolor{purple}{\dfrac{\sqrt{3}}{2}}$, putting this into the equation we can solve for $\textcolor{limegreen}{y}$.

$\textcolor{purple}{\dfrac{\sqrt{3}}{2}} = \dfrac{\textcolor{limegreen}{y}}{\textcolor{blue}{10}}$

$\dfrac{\textcolor{blue}{10}\textcolor{purple}{\sqrt{3}}}{\textcolor{purple}{2}} = \textcolor{limegreen}{y}$

$5\sqrt{3} = \textcolor{limegreen}{y}$

Level 4-5 GCSE    ## Example Questions

Each term is one of a small list of common trigonometry values that students are required to remember. Here,

$\sin(60\degree)=\dfrac{\sqrt{3}}{2}$

as well as,

$\cos(30\degree)=\dfrac{\sqrt{3}}{2}$

Thus the calculation is,

$\sin(60\degree)+ \cos(30\degree)=\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}=\sqrt{3}$

Each term is one of a small list of common trigonometry values that students are required to remember:

$\tan(45\degree)=1$

$\sin(30\degree)=\dfrac{1}{2}$

and

$\tan(60\degree)=\sqrt{3}$

Thus the calculation is,

$\dfrac{\tan(45\degree)}{\sin(30\degree)} \times 10\tan(60\degree)=\dfrac{1}{\frac{1}{2}} \times 10(\sqrt{3})=2\times10\sqrt{3}=20\sqrt{3}$

Each term is one of a small list of common trigonometry values that students are required to remember:

$\tan(45\degree)=1$

$\cos(30\degree)=\dfrac{\sqrt{3}}{2}$

and

$\cos(60\degree)=\dfrac{1}{2}$

Thus the calculation is,

$\dfrac{\tan(45\degree)+\cos{(30\degree)}}{\tan(45\degree)} \times \cos(60\degree)=\dfrac{1+\frac{\sqrt{3}}{2}}{1} \times \dfrac{1}{2}=\dfrac{2+\sqrt{3}}{2}\times \dfrac{1}{2}=\dfrac{2+\sqrt{3}}{4}$

This question requires the use of Pythagoras. We are given the hypotenuse and want the find the opposite side length, hence,

$\sin(30\degree)=\dfrac{x}{16}$

Thus to find $x$ we simply multiply both sides by 16 and use the common trigonometry value of $\sin(30\degree) =\dfrac{1}{2}$

$x=16\times\sin(30\degree)=16\times \dfrac{1}{2}=8$ cm.

This question requires the use of Pythagoras. We are given the adjacent length and want the find the opposite side length, hence,

$\tan(30\degree)=\dfrac{x}{12}$

Thus to find $x$ we multiply both sides by $12$ and use the common trigonometry value of $\tan(30\degree) =\dfrac{1}{\sqrt{3}}$

$x=12\times\tan(30\degree)=12\times \dfrac{1}{\sqrt{3}}=12\times \dfrac{\sqrt{3}}{3}=4\sqrt{3}$ cm.

Level 6-7GCSEKS3

## Worksheet and Example Questions

### (NEW) Trig Common Values Exam Style Questions - MME

Level 6-7 GCSENewOfficial MME

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