The R Formula
Introduction
The aim of this article is to help you understand the R Formula and to allow you to be able to convert expressions such as a \sin \theta+b \cos \theta to the form r \cos (\theta \pm \alpha) \operatorname{or} r \sin (\theta \pm \alpha)
R Formula - Introduction
A modification of the compound angle formula allows you to simplify expressions such as 4 \sin \theta+3 \cos \theta and to solve equations.
In order to find a single expression for 4 \sin \theta+3 \cos \theta his is matched to the compound angle formula r \sin (\theta+\alpha) because this has sinθ in the first term and cosθ in the second term as well as a plus sign.
\begin{gathered} 4 \sin \theta+3 \cos \theta-r \sin (\theta+\alpha) \\ 4 \sin \theta+3 \cos \theta \equiv r(\sin \theta \cos \alpha+\cos \theta \sin \alpha) \end{gathered}Comparing the coefficients of sinθ and cosθ:
\begin{aligned} & 4=r \cos \alpha \\ & 3=r \sin \alpha \end{aligned}The value of r and \theta can be obtained by squaring and adding these equations as shown:
\begin{gathered} 3^2+4^2=r^2 \cos ^2 \alpha+r^2 \sin ^2 \alpha \\ 25=r^2\left(\cos ^2 \alpha+\sin ^2 \alpha\right) \\ 25=r^2 \\ \therefore r=5 \end{gathered}The value of \alpha an be obtained by dividing these equations as shown:
\frac{3}{4}=\frac{r \sin \alpha}{r \cos \alpha}\tan \alpha=\frac{3}{4} \therefore \alpha=36.9^{\circ}This means that the original expression can be written as;
4 \sin \theta+3 \cos \theta=5 \sin \left(\theta+36.9^{\circ}\right)
R Formula Example 1
Solution
Part a)
\begin{aligned} & r \sin (\theta-\alpha)=r(\sin \theta \cos \alpha-\cos \theta \sin \alpha) \\ & \qquad \sqrt{3} \sin \theta-\cos \theta=r(\sin \theta \cos \alpha-\cos \theta \sin \alpha) \end{aligned}Comparing coefficients of sinθ and cosθ:
\begin{gathered} \sqrt{3}=r \cos \alpha \\ 1=r \sin \alpha \end{gathered}Squaring and adding these equations gives:
\begin{gathered} 3+1=r^2 \\ \therefore r=2 \end{gathered}Dividing these two equations gives:
\begin{gathered} \tan \alpha=\frac{1}{\sqrt{3}} \therefore \alpha=\frac{\pi}{6} \\ \sqrt{3} \sin \theta-\cos \theta=2 \sin \left(\theta-\frac{\pi}{6}\right) \end{gathered}Part b)
The maximum and minimum values of y=\sin x \text { is } \pm 1
Now y=2 \sin \left(\theta-\frac{\pi}{6}\right) represents a stretch of scale factor 2 parallel to the y axis.
So the maximum and minimum values are ±2
Part c)
To solve \sqrt{3} \sin \theta-\cos \theta=1
This can be written as:
\begin{gathered} 2 \sin \left(\theta-\frac{\pi}{6}\right)=1 \\ \sin \left(\theta-\frac{\pi}{6}\right)=\frac{1}{2} \\ \theta-\frac{\pi}{6}=\frac{\pi}{6} \end{gathered}OR
\begin{gathered} \theta-\frac{\pi}{6}=\frac{5 \pi}{6} \\ \therefore \theta=\frac{\pi}{3}, \pi \end{gathered}
R Formula – Example 2
Solution
\begin{gathered} 3 \sin x+4 \cos x=r \sin (x+\alpha) \\ 3 \sin x+4 \cos x=r(\sin x \cos \alpha+\cos x \sin \alpha) \end{gathered}Comparing coefficients of sinx and cosx:
\begin{aligned} & 3=r \cos \alpha \\ & 4=r \sin \alpha \end{aligned}Squaring and adding these equations gives:
3^2+4^2=r^2 \therefore r=5Dividing these equations gives:
\begin{gathered} \tan \alpha=\frac{4}{3} \therefore \alpha=53.1^{\circ} \\ 3 \sin x+4 \cos x=5 \sin \left(x+53.1^{\circ}\right) \end{gathered}Can now solve:
\begin{gathered} 5 \sin \left(x+53.1^{\circ}\right)-3 \\ \sin \left(x+53.1^{\circ}\right)=0.6 \\ x+53.1^{\circ}=36.9^{\circ}, 143.1^{\circ} \\ x=-16.2^{\circ}, 90^{\circ} \\ \text { i.e. } \\ x=90^{\circ}, 343^{\circ} \end{gathered}Here are some R Formula Exam Style Questions
Question 1
Part a)
To do this part of the question you need to first expand using the compound angle formula and to then compare coefficients in order to obtain values for R and \alpha.
\begin{aligned} f(x) & =3 \cos x-4 \sin x \\ \cos (A+B) & =\cos A \cos B-\sin A \sin B \\ R \cos (x+\alpha) & =R \cos x \cos \alpha-R \sin x \sin \alpha \end{aligned}3=R \cos \alpha \quad 4=R \sin \alpha\begin{aligned} R & =\sqrt{3^2+4^2} & \tan \alpha & =\frac{4}{3} \\ & =5 & \alpha & =53.1 \end{aligned}Part b)
Use your answer to part a) to help with this part of the question. Remember to be looking at the range and also if working in degrees or radians.
\begin{aligned} 5 \cos (x+53.1) & =1 \\ \cos (x+53.1) & =1 / 5 \\ x+53.1 & =78.5,281.5 \\ x & =25.3,228.4 \end{aligned}Part c)
Your answer to part a) is no more than a stretch and translation. Remember that the curve of y = cosx is valid between 1 so the minimum value of y = cosx. This then means that the minimum value of y=5 \cos (x+53.1) \text { is }-5
Part d)
Here you are determining the angle for which the minimum occurs. Because we know that the minimum occurs at -5 it is a case of solving,
-5=5 \cos (x+53.1)\begin{aligned} 5 \cos (x+53.1) & =-5 \\ \cos (x+53.1) & =-1 \\ x+53.1 & =180 \\ x & =126.9^{\circ} \end{aligned}Question 2
Part a)
Perform the compound angle expansion and compare coefficients. These are quite standard questions and the process is also quite standard. Make sure that you are able to do it quite quickly as it is easy marks to gain.
$5 \sin x+12 \cos x$ $R \sin (x+\alpha)=R \sin x \cos \alpha+R \cos x \sin \alpha$5=R \cos \alpha \quad 12=R \sin \alpha\begin{aligned} R & =\sqrt{5^2+12^2} & \tan \alpha & =12 / 5 \\ & =13 & \alpha & =67.4 \end{aligned}13 \sin (x+67.4)Part b)
Remember that your answer to part a) is simply a stretch and translation of the curve y = sinx. The maximum and minimum values of y = sinx are 1. So the maximum value is 13.
You can then find the angle for which this occurs as follows:
\begin{aligned} 13 \sin (x+67.4) & =13 \\ \sin (x+67.4) & =1 \\ x+67.4 & =90 \\ x & =22.6 \end{aligned}You can get more complicated R Formula questions as we will create a new blog shortly looking at more longer questions. But even in those, you are still doing what has been done here.
Make sure that you are comfortable with the methods discussed here and look out for future blog posts.