Trigonometry Common Values Revision and Worksheets | Maths Made Easy

Trigonometry Common Values Revision and Worksheets

Level 6-7

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

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.

Trigonometry Values to Learn
Level 4-5

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}

Trigonometry Question Image (CAH)
Level 4-5

GCSE Maths Revision Cards

(242 Reviews) £8.99
View Buy this product on Amazon

Take an Online Exam

Trigonometry Common Values Online Exam

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.

Worksheets and Exam Questions

MME

(NEW) Trig Common Values Exam Style Questions - MME

Level 6-7 New Official MME