A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

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.

A Level   ## 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 $\textcolor{maroon}{R}\textcolor{limegreen}{\cos} (\textcolor{purple}{\theta} \mp \textcolor{orange}{\alpha})$.

A Level   ## 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.

A Level   ## Factor Formulae

We’ve also got four new formulae…

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

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

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

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

A Level   A Level   ## 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

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

A Level   ## 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 $8\textcolor{blue}{\sin} x + 6\textcolor{limegreen}{\cos} x = 10\textcolor{limegreen}{\cos} (x - 53.13°)$.

A Level   ## Example Questions

Let $a = 5 = R\cos \alpha$ and $b = 3 = R\sin \alpha$.

Then,

$R = \sqrt{5^2 + 3^2} = \sqrt{34}$ and $\alpha = \tan ^{-1} \dfrac{3}{5} = 30.96°$

Giving

$5\cos x - 3\sin x = \sqrt{34}\cos (x + 30.96°)$

$6\sin \left( \dfrac{\pi}{4}\right) \sin \left( \dfrac{3\pi}{2} \right)$

$= 3 \times 2\sin \left( \dfrac{3\pi}{2}\right) \sin \left( \dfrac{\pi}{4} \right)$

$= 3\left( \cos \dfrac{5\pi}{4} - \cos \dfrac{7\pi}{4}\right)$

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

We have $a = 3$ and $b = 4$.

Then, we have $R^2 = a^2 + b^2$, or, $R = \sqrt{a^2 + b^2} = \sqrt{25} = 5$.

Therefore, the minimum and maximum values of $3\cos x + 4\sin x$ must be $-5$ and $5$.

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