# Addition and Double Angle Formulae

A LevelAQAEdexcelOCR

## 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.

A LevelAQAEdexcelOCR

## 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 LevelAQAEdexcelOCR

## Note:

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

$\textcolor{limegreen}{\cos} (\textcolor{purple}{A} - \textcolor{orange}{B}) = \textcolor{limegreen}{\cos} \textcolor{purple}{A} \textcolor{limegreen}{\cos} \textcolor{orange}{B} + \textcolor{blue}{\sin} \textcolor{purple}{A} \textcolor{blue}{\sin} \textcolor{orange}{B}$

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## 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 LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

## 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 LevelAQAEdexcelOCR

## Addition and Double Angle Formulae Example Questions

$\cos 165° = \cos (210° - 45°)$

$= \cos 210° \cos 45° + \sin 210° \sin 45°$

$= \left( \dfrac{-\sqrt{3}}{2} \times \dfrac{1}{\sqrt{2}}\right) + \left( \dfrac{-1}{2} \times \dfrac{1}{\sqrt{2}}\right)$

$= \dfrac{-\sqrt{3} - 1}{2\sqrt{2}}$

Gold Standard Education

$\tan 150° = \dfrac{2(2 + \sqrt{3})}{1 - (2 + \sqrt{3})^2}$

$= \dfrac{4 + 2\sqrt{3}}{-6 - 4\sqrt{3}} = \dfrac{2 + \sqrt{3}}{-3 - 2\sqrt{3}}$

$= \dfrac{(2 + \sqrt{3})(-3 + 2\sqrt{3})}{(-3 - 2\sqrt{3})(-3 + 2\sqrt{3})}$

$= \dfrac{-6 - 3\sqrt{3} + 4\sqrt{3} + 6}{9 - 12}$

$= \dfrac{\sqrt{3}}{-3} = \dfrac{-1}{\sqrt{3}}$

Gold Standard Education

$\sin \left( x + \dfrac{\pi}{2}\right) = \sin x \cos \dfrac{\pi}{2} + \sin \dfrac{\pi}{2} \cos x$

$= (\sin x \times 0) + (\cos x \times 1)$

$= \cos x$

Gold Standard Education

A Level

A Level

## Addition and Double Angle Formulae Worksheet and Example Questions

### Double Angle Formulae

A Level

Gold Standard Education

A Level