Trigonometric Equations - R method

The $R$ Method is a technique used to solve trigonometric equations of the form

$$a\sin x \pm b\cos x = c$$

by applying the addition formulae.

Express $3\sin x + 4 \cos x$ in the form $R\sin (x+\alpha)$.

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Our goal is to solve for $R$ and $\alpha$ in the following equation:

$$\begin{aligned} 3\sin x + 4 \cos x &\equiv R\sin (x+\alpha) \\ & = R\sin x \cos \alpha + R \cos x \sin \alpha\\\\ \Rightarrow R\cos \alpha &= 3,\quad R\sin \alpha = 4. \end{aligned}$$

Taking their ratio leads to

$$\displaystyle \tan \alpha = \frac{4}{3}\implies \alpha = \arctan \frac{4}{3} = 53^\circ 8'.$$

Substituting the value of $\alpha$ into the previous equation gives

$$R\cos53^\circ 8' = 3\implies R = 5.$$

Finally,

$$3\sin x + 4 \cos x \equiv 5\sin (x+53^\circ 8').\ _\square$$

Taking the most suitable form of trigonometric addition formula will save you a lot of work. The equations below state the best $R$ method in different cases.

$$a\sin x + b \cos x \equiv R\sin (x+\alpha)$$ $$a\sin x - b \cos x \equiv R\sin (x-\alpha)$$ $$a\cos x + b \sin x \equiv R\cos (x-\alpha)$$ $$a\cos x - b \sin x \equiv R\cos (x+\alpha)$$

$R$ method can be very effective when solving trigonometric equations and inequalities.

Solve the equation $3\sin x + 4 \cos x = 1$ for $0\leq x \leq 180^\circ.$

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Applying the $R$ method, we have

$$\begin{aligned} 3 \sin x + 4 \cos x &= 5\sin (x+53^\circ 8') \\ 5\sin (x+53^\circ 8') &= 1 \\ \sin (x+53^\circ 8') &= \frac{1}{5} \\ x+53^\circ 8' &= 168^\circ 28' \\ x &= 115^\circ 20'.\ _\square \end{aligned}$$

What is the maximum value of $3\sin x + 4 \cos x?$

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Since the maximum value of the sine function is $1$, we have

$$\begin{aligned} 3 \sin x + 4 \cos x &= 5\sin (x+53^\circ 8') \\ &\leq 5.\ _\square \end{aligned}$$