# Addition and Double Angle Formulae

## Addition and Double Angle Formulae Revision

**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}}

**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}

**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.

**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})

## Addition and Double Angle Formulae Example 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