R Addition Formulae

We use the R formulae when we have a set of $\textcolor{blue}{\sin}$ and $\textcolor{limegreen}{\cos}$ functions which we want to collect as a single trig function. Essentially, this is the Addition and Double Angle Formulae section, in reverse.

Make sure you are happy with the following topics before continuing.

Addition and Double Angle Formulae

Origins of the R Formulae

Don’t worry, we’re not teaching you another equation – we’re teaching you a new trick for an old equation.

In the Addition and Double Angle Formulae section, we claimed that we could split $\textcolor{blue}{\sin} (\textcolor{purple}{A} \pm \textcolor{orange}{B})$ and $\textcolor{limegreen}{\cos}(\textcolor{purple}{A} \pm \textcolor{orange}{B})$ into separate $\textcolor{blue}{\sin} \textcolor{purple}{A}, \textcolor{blue}{\sin} \textcolor{orange}{B}, \textcolor{limegreen}{\cos} \textcolor{purple}{A}$ and $\textcolor{limegreen}{\cos} \textcolor{orange}{B}$ terms.

We’re now going to look at how to collect a linear combination of those terms into one expression, $\textcolor{maroon}{R}\textcolor{blue}{\sin} (\textcolor{purple}{\theta} \pm \textcolor{orange}{\alpha})$ or $\textcolor{maroon}{R}\textcolor{limegreen}{\cos} (\textcolor{purple}{\theta} \mp \textcolor{orange}{\alpha})$ .

Using R Formulae

So, let’s say we have a general expression,

$a\textcolor{blue}{\sin} \textcolor{purple}{\theta} \pm b\textcolor{limegreen}{\cos} \textcolor{purple}{\theta}$

Surely we can express that as

$\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha} \textcolor{blue}{\sin} \textcolor{purple}{\theta} \pm \textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha} \textcolor{limegreen}{\cos} \textcolor{purple}{\theta}$

where $a = \textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha}$ and $b = \textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha}$ , right?

$\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha} \textcolor{blue}{\sin} \textcolor{purple}{\theta} \pm \textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha} \textcolor{limegreen}{\cos} \textcolor{purple}{\theta} = \textcolor{maroon}{R}(\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha} \textcolor{blue}{\sin} \textcolor{purple}{\theta} \pm \textcolor{blue}{\sin} \textcolor{orange}{\alpha} \textcolor{limegreen}{\cos} \textcolor{purple}{\theta}) = \textcolor{maroon}{R}\textcolor{blue}{\sin} (\textcolor{purple}{\theta} \pm \textcolor{orange}{\alpha})$

Then, we can find a value of $\textcolor{orange}{\alpha}$ , or rather, $\textcolor{red}{\tan} \textcolor{orange}{\alpha}$ , by finding $\dfrac{b}{a} = \dfrac{\textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha}}{\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha}} = \textcolor{red}{\tan} \textcolor{orange}{\alpha}$ , so $\textcolor{orange}{\alpha} = \tan ^{-1}\dfrac{b}{a}$ .

We can also find a value of $\textcolor{maroon}{R}$ , by doing the calculation $a^2 + b^2 = \textcolor{maroon}{R}^2\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{maroon}{R}^2\textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha} = \textcolor{maroon}{R}^2(\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha}) = \textcolor{maroon}{R}^2$ , so $\textcolor{maroon}{R} = \sqrt{a^2 + b^2}$ .

Substituting our values of $\textcolor{orange}{\alpha}$ and $\textcolor{maroon}{R}$ gives us an expression for $a\textcolor{blue}{\sin} \textcolor{purple}{\theta} \pm b\textcolor{limegreen}{\cos} \textcolor{purple}{\theta}$ in the form $\textcolor{maroon}{R}\textcolor{blue}{\sin} (\textcolor{purple}{\theta} \pm \textcolor{orange}{\alpha})$ .

We can do a similar process to find an expression in $\color{maroon}R\color{limegreen}\cos \color{grey}(\color{purple}\theta \color{grey}\mp \color{orange}\alpha\color{grey})$ , also.

Factor Formulae

We’ve also got four new formulae…

$\textcolor{blue}{\sin} \textcolor{purple}{A} + \textcolor{blue}{\sin} \textcolor{orange}{B} = 2\textcolor{blue}{\sin} \left( \dfrac{\textcolor{purple}{A} + \textcolor{orange}{B}}{2}\right) \textcolor{limegreen}{\cos} \left( \dfrac{\textcolor{purple}{A} - \textcolor{orange}{B}}{2}\right)$

$\textcolor{blue}{\sin} \textcolor{purple}{A} - \textcolor{blue}{\sin} \textcolor{orange}{B} = 2\textcolor{limegreen}{\cos} \left( \dfrac{\textcolor{purple}{A} + \textcolor{orange}{B}}{2}\right) \textcolor{blue}{\sin} \left( \dfrac{\textcolor{purple}{A} - \textcolor{orange}{B}}{2}\right)$

$\textcolor{limegreen}{\cos} \textcolor{purple}{A} + \textcolor{limegreen}{\cos} \textcolor{orange}{B} = 2\textcolor{limegreen}{\cos} \left( \dfrac{\textcolor{purple}{A} + \textcolor{orange}{B}}{2}\right) \textcolor{limegreen}{\cos} \left( \dfrac{\textcolor{purple}{A} - \textcolor{orange}{B}}{2}\right)$

$\textcolor{limegreen}{\cos} \textcolor{purple}{A} - \textcolor{limegreen}{\cos} \textcolor{orange}{B} = 2\textcolor{blue}{\sin} \left( \dfrac{\textcolor{purple}{A} + \textcolor{orange}{B}}{2}\right) \textcolor{blue}{\sin} \left( \dfrac{\textcolor{purple}{A} - \textcolor{orange}{B}}{2}\right)$

Example 1: Converting With R Formulae

That might’ve been a lot of information at once, so let’s try an example.

Express $8\textcolor{blue}{\sin} x + 6\textcolor{limegreen}{\cos} x$ in the form $\textcolor{maroon}{R}\textcolor{blue}{\sin} (x + \textcolor{orange}{\alpha})$ , where $0° \leq \textcolor{orange}{\alpha} \leq 90°$ .

[4 marks]

Let $\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha} = 8$ , and $\textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha} = 6$ .

So, then, $\textcolor{maroon}{R}^2 = \textcolor{maroon}{R}^2(\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha}) = \textcolor{maroon}{R}^2\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{maroon}{R}^2\textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha} = 8^2 + 6^2 = 100$ , giving $\textcolor{maroon}{R} = 10$ .

Also, $\textcolor{orange}{\alpha} = \tan ^{-1}(\textcolor{red}{\tan} \textcolor{orange}{\alpha}) = \tan ^{-1} \left( \dfrac{\textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha}}{\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha}}\right) = \tan ^{-1} \dfrac{6}{8} = 36.87°$ .

So, to summarise, we have

$8\textcolor{blue}{\sin} x + 6\textcolor{limegreen}{\cos} x = 10\textcolor{blue}{\sin} (x + 36.87°)$

Example 2: Converting With R Formulae (Again)

Express $8\textcolor{blue}{\sin} x + 6\textcolor{limegreen}{\cos} x$ in the form $\textcolor{maroon}{R}\textcolor{limegreen}{\cos} (x - \textcolor{orange}{\alpha})$ , where $0° \leq \textcolor{orange}{\alpha} \leq 90°$ .

[4 marks]

Let $\textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha} = 8$ , and $\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha} = 6$ .

So, then, $\textcolor{maroon}{R}^2 = \textcolor{maroon}{R}^2(\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha}) = \textcolor{maroon}{R}^2\textcolor{limegreen}{\cos} ^2 \textcolor{orange}{\alpha} + \textcolor{maroon}{R}^2\textcolor{blue}{\sin} ^2 \textcolor{orange}{\alpha} = 6^2 + 8^2 = 100$ , giving $\textcolor{maroon}{R} = 10$ .

Also, $\textcolor{orange}{\alpha} = \tan ^{-1}(\textcolor{red}{\tan} \textcolor{orange}{\alpha}) = \tan ^{-1} \left( \dfrac{\textcolor{maroon}{R}\textcolor{blue}{\sin} \textcolor{orange}{\alpha}}{\textcolor{maroon}{R}\textcolor{limegreen}{\cos} \textcolor{orange}{\alpha}}\right) = \tan ^{-1} \dfrac{8}{6} = 53.13°$ .

So, to summarise, we have $8\textcolor{blue}{\sin} x + 6\textcolor{limegreen}{\cos} x = 10\textcolor{limegreen}{\cos} (x - 53.13°)$ .