Trigonometry Common Values Revision and Worksheets

Trigonometry Common Values Revision and Worksheets

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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:

Trigonometry

Level 6-7 GCSE AQA Edexcel OCR WJEC
Trigonometry Values to Learn

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

Trigonometry Values to Learn

You need to remember all of these values.

Level 4-5 GCSE AQA Edexcel OCR WJEC
Trigonometry Question Image (CAH)

Example

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

[2 marks]

Trigonometry Question Image (CAH)

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 AQA Edexcel OCR WJEC

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.

Related Topics

MME

Trigonometry

Level 6-7GCSEKS3

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