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).

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:

• The sine form:
  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 $\alpha=\arctan\frac{b}{a}.$

• The cosine form:
  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 $\alpha=\arctan\frac{a}{b}.$

First, let's prove the sine form. Let $a=R\cos\alpha$ and $b=R\sin\alpha.$ Then the following equations correspond:

$$\begin{aligned} a^2+b^2&=R^2(\cos^2\alpha+\sin^2\alpha)\\&=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}\\&=\frac{b}{a}\\ a\sin\theta+b\cos\theta&=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha\\&=R\sin(\theta+\alpha).\ _\square \end{aligned}$$

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To prove the cosine form, let $a=R\sin\alpha$ and $b=R\cos\alpha.$ Then the following equations correspond:

$$\begin{aligned} a^2+b^2&=R^2(\sin^2\alpha+\cos^2\alpha)=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}=\frac{a}{b}\\ a\sin\theta+b\cos\theta&=R\cos\theta\cos\alpha+R\sin\theta\sin\alpha=R\cos(\theta-\alpha).\ _\square \end{aligned}$$

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

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

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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.

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(1) Using the sine form:
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 cosine form:
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 $x$:

$$\sqrt{3}\sin x-\cos x=\sqrt{2},\quad -\pi<x<\pi.$$

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

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

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

$$\begin{aligned} x-\frac{\pi}{6}=\frac{\pi}{4}&\implies x=\frac{5}{12}\pi\\ x-\frac{\pi}{6}=\frac{3}{4}\pi&\implies x=\frac{11}{12}\pi. \end{aligned}$$

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.

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 $x,$ 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

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

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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 $-3+7=4.$ $_\square$

• Trigonometry