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 $5$.

A Level

A Level

A Level

A Level

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99

Transition 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!

£8.99