# Trigonometric R method

The **trigonometric R method** is a method of rewriting a weighted sum of sines and cosines as a single instance of sine (or cosine). This allows for easier analysis in many cases, as a single instance of a basic trigonometric function is often easier to work with than multiple are.

The R method is most often used to find the **extrema** (maximum and minimum) of combinations of trigonometric functions, since the extrema of a basic trigonometric function are easy to work with (both sine and cosine have a minimum of -1 and a maximum of 1).

## Formal Statement and Proof

The R method can be used to convert an expression of the form \(a\sin\theta+b\cos\theta\) into a single instance of either sine or cosine:

(1) The

sineform:

For any \(a,b,\theta\in\mathbb{R},\) the following corresponds: \[a\sin\theta+b\cos\theta=R\sin(\theta+\alpha),\] where \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\frac{b}{a},\) or \(\alpha=\arctan\frac{b}{a}.\) \(_\square\)(2) The

cosineform:

For any \(a,b,\theta\in\mathbb{R},\) the following corresponds: \[a\sin\theta+b\cos\theta=R\cos(\theta-\alpha),\] where \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\frac{a}{b},\) or \(\alpha=\arctan\frac{a}{b}.\) \(_\square\)

First, let's prove the

sineform. Let \(a=R\cos\alpha\) and \(b=R\sin\alpha.\) Then the following equations correspond:\[\begin{align} a^2+b^2&=R^2(\cos^2\alpha+\sin^2\alpha)\\&=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}\\&=\frac{b}{a}\\ a\sin\theta+b\cos\theta&=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha\\&=R\sin(\theta+\alpha).\ _\square \end{align}\]

To prove the

cosineform, let \(a=R\sin\alpha\) and \(b=R\cos\alpha.\) Then the following equations correspond:\[\begin{align} a^2+b^2&=R^2(\sin^2\alpha+\cos^2\alpha)=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}=\frac{a}{b}\\ a\sin\theta+b\cos\theta&=R\cos\theta\cos\alpha+R\sin\theta\sin\alpha=R\cos(\theta-\alpha).\ _\square \end{align}\]

## Using the R Method

## Given the condition \(R>0\) and \(0<\theta<\frac{\pi}{2},\) what are the appropriate values of \(R\) and \(\theta\) for the following equation:

\[\sin\frac{\pi}{8}+\cos\frac{\pi}{8}=R\sin\theta?\]

Since \(R^2=1^2+1^2=2\) and \(\arctan\frac{1}{1}=\frac{\pi}{4},\) we have

\[\sin\frac{\pi}{8}+\cos\frac{\pi}{8}=\sqrt{2}\sin\left(\frac{\pi}{8}+\frac{\pi}{4}\right)=\sqrt{2}\sin\frac{3}{8}\pi.\ _\square\]

Therefore, \(R=\sqrt{2}\) and \(\theta=\frac{3}{8}\pi.\ _\square\)

## Simplify \(\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}\) using the R method.

(1) Using the

sineform:

Since \(R^2=1^2+(\sqrt{3})^2=4\) and \(\arctan\sqrt{3}=\frac{\pi}{3},\) we have \[\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}=2\sin\left(\frac{\pi}{6}+\frac{\pi}{3}\right)=2\sin\frac{\pi}{2}=2.\ _\square\](2) Using the

cosineform:

Since \(R^2=1^2+(\sqrt{3})^2=4\) and \(\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{6},\) we have \[\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}=2\cos\left(\frac{\pi}{6}-\frac{\pi}{6}\right)=2\cos0=2.\ _\square\]

## Solve for \(x\):

\[\sqrt{3}\sin x-\cos x=\sqrt{2}\ \ \ (-\pi<x<\pi).\]

Since \(R^2=(\sqrt{3})^2+(-1)^2=4\) and \(\arctan\frac{-1}{\sqrt{3}}=-\frac{\pi}{6},\) we have

\[\begin{align} \sqrt{3}\sin x-\cos x=2\sin\left(x-\frac{\pi}{6}\right)&=\sqrt{2}\\ \Rightarrow \sin\left(x-\frac{\pi}{6}\right)&=\frac{\sqrt{2}}{2}. \end{align}\]

Let \(x-\frac{\pi}{6}=\theta.\) Given \(-\pi<x<\pi,\) we have \(-\frac{7}{6}\pi<\theta<\frac{5}{6}\pi.\) In this interval the values of \(\theta\) that satisfy \(\sin\theta=\frac{\sqrt{2}}{2}\) are \(\frac{\pi}{4}\) and \(\frac{3}{4}\pi.\) Thus we have

\[\begin{align} x-\frac{\pi}{6}=\frac{\pi}{4}&\Rightarrow x=\frac{5}{12}\pi\\ x-\frac{\pi}{6}=\frac{3}{4}\pi&\Rightarrow x=\frac{11}{12}\pi. \end{align}\]

Therefore the answer is \(x=\frac{5}{12}\pi, \frac{11}{12}\pi.\) \(_\square\)

When using the R method, there are numerous possible values of \(R\) and \(\alpha,\) since sine and cosine are periodic functions. Adding \(2n\pi\) to \(\alpha\) for any integer \(n\) would give the same answer. Also, adding \((2n-1)\pi\) to \(\alpha\) for any integer \(n\) and changing the sign of \(R\) would also be equivalent to the answer. However we most often use \(-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\) and \(R>0\) for convenience.

## Applications to Extrema

The R method is frequently used to find the maximum or minimum of equations in the form of \(a\sin\theta+b\cos\theta.\) It is known that \(-1\leq\sin x\leq1\) and \(-1\leq\cos x\leq1\) for any real number \(x,\) and so the maximum and minimum of \(a\sin\theta+b\cos\theta\) are \(R\) and \(-R,\) respectively.

## Find the sum of the maximum and minimum values of

\[3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)+2.\]

Since \(3^2+(-4)^2=5^2,\) we have \(R=5.\) This implies that

\[-5\leq3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)\leq5\\ \Rightarrow -3\leq3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)+2\leq7.\]

Therefore the answer is \(-3+7=4.\) \(_\square\)

## See Also

**Cite as:**Trigonometric R method.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/trigonometric-r-method/