Addition and Double Angle Formulae
Addition and Double Angle Formulae
Addition and Double Angle Formulae
We’re now about to take a look at some formulae which describe angle addition.
If you don’t know your key trig values already, now would be the time to learn!
Make sure you are happy with the following topics before continuing.
Finding Expressions for Addition Formulae
Here’s three new formulae in \textcolor{blue}{\sin}, \textcolor{limegreen}{\cos} and \textcolor{red}{\tan}:
\textcolor{blue}{\sin} (\textcolor{purple}{A} ± \textcolor{orange}{B}) = \textcolor{blue}{\sin} \textcolor{purple}{A} \textcolor{limegreen}{\cos} \textcolor{orange}{B} ± \textcolor{blue}{\sin} \textcolor{orange}{B} \textcolor{limegreen}{\cos} \textcolor{purple}{A}
\textcolor{limegreen}{\cos} (\textcolor{purple}{A} ± \textcolor{orange}{B}) = \textcolor{limegreen}{\cos} \textcolor{purple}{A} \textcolor{limegreen}{\cos} \textcolor{orange}{B} \mp \textcolor{blue}{\sin} \textcolor{purple}{A} \textcolor{blue}{\sin} \textcolor{orange}{B}
\textcolor{red}{\tan} (\textcolor{purple}{A} ± \textcolor{orange}{B}) = \dfrac{\textcolor{red}{\tan} \textcolor{purple}{A} ± \textcolor{red}{\tan} \textcolor{orange}{B}}{1 \mp \textcolor{red}{\tan} \textcolor{purple}{A} \textcolor{red}{\tan} \textcolor{orange}{B}}
A LevelNote:
You might have noticed the “minus-plus” symbols above (\mp). This is no mistake, and it is not the same as “plus-minus, \pm“. The important thing to remember with this notation is that whichever symbol is chosen (top or bottom), must be used on the other side of the equation.
So, for example,
\textcolor{limegreen}{\cos} (\textcolor{purple}{A} + \textcolor{orange}{B}) = \textcolor{limegreen}{\cos} \textcolor{purple}{A} \textcolor{limegreen}{\cos} \textcolor{orange}{B} - \textcolor{blue}{\sin} \textcolor{purple}{A} \textcolor{blue}{\sin} \textcolor{orange}{B}
and
\textcolor{limegreen}{\cos} (\textcolor{purple}{A} - \textcolor{orange}{B}) = \textcolor{limegreen}{\cos} \textcolor{purple}{A} \textcolor{limegreen}{\cos} \textcolor{orange}{B} + \textcolor{blue}{\sin} \textcolor{purple}{A} \textcolor{blue}{\sin} \textcolor{orange}{B}
Double Angle Formulae
We can extend our addition formulae to two equal angles, also.
So, we have
\textcolor{blue}{\sin (2A)} = 2\textcolor{blue}{\sin A} \textcolor{limegreen}{\cos A}
\begin{aligned}\textcolor{limegreen}{\cos (2A)} &= \textcolor{limegreen}{\cos ^2 A} - \textcolor{blue}{\sin ^2 A}\\[1.2em]&=2\textcolor{limegreen}{\cos^2 A}-1\\[1.2em]&=1-2\textcolor{blue}{\sin^2 A}\end{aligned}
\textcolor{red}{\tan (2A)} = \dfrac{2\textcolor{red}{\tan A}}{1 - \textcolor{red}{\tan ^2 A}}
No worries if you forget these, you can just derive them from the addition formulae by setting B = A.
A LevelTransition Maths Cards
The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!
Example: Finding Exact Values
Find the exact value of \textcolor{blue}{\sin} 75°, in the form \dfrac{1}{a\sqrt{b}}(c + \sqrt{d}).
[3 marks]
\textcolor{blue}{\sin} 75° = \textcolor{blue}{\sin} (30° + 45°)
= \textcolor{blue}{\sin} 30° \textcolor{limegreen}{\cos} 45° + \textcolor{blue}{\sin} 45° \textcolor{limegreen}{\cos} 30°
= \left( \dfrac{1}{2} \times \dfrac{1}{\sqrt{2}} \right) + \left( \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} \right)
= \dfrac{1}{2\sqrt{2}} + \dfrac{\sqrt{3}}{2\sqrt{2}} = \dfrac{1}{2\sqrt{2}}(1 + \sqrt{3})
A LevelExample Questions
Question 1: Find the exact value of \cos 165°.
[3 marks]
\cos 165° = \cos (210° - 45°)
= \cos 210° \cos 45° + \sin 210° \sin 45°
= \left( \dfrac{-\sqrt{3}}{2} \times \dfrac{1}{\sqrt{2}}\right) + \left( \dfrac{-1}{2} \times \dfrac{1}{\sqrt{2}}\right)
= \dfrac{-\sqrt{3} - 1}{2\sqrt{2}}
Question 2: Given that \tan 75° = 2 + \sqrt{3}, find the exact value of \tan 150°.
[2 marks]
\tan 150° = \dfrac{2(2 + \sqrt{3})}{1 - (2 + \sqrt{3})^2}
= \dfrac{4 + 2\sqrt{3}}{-6 - 4\sqrt{3}} = \dfrac{2 + \sqrt{3}}{-3 - 2\sqrt{3}}
= \dfrac{(2 + \sqrt{3})(-3 + 2\sqrt{3})}{(-3 - 2\sqrt{3})(-3 + 2\sqrt{3})}
= \dfrac{-6 - 3\sqrt{3} + 4\sqrt{3} + 6}{9 - 12}
= \dfrac{\sqrt{3}}{-3} = \dfrac{-1}{\sqrt{3}}
Question 3: Using angle addition formulae, prove that \sin \left( x + \dfrac{\pi}{2}\right) = \cos x.
[2 marks]
\sin \left( x + \dfrac{\pi}{2}\right) = \sin x \cos \dfrac{\pi}{2} + \sin \dfrac{\pi}{2} \cos x
= (\sin x \times 0) + (\cos x \times 1)
= \cos x
Related Topics
Basic Trig Identities
A LevelWorksheet and Example Questions
Double Angle Formulae
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