Simplifying $a \cos x + b \sin x$

In a nutshell

You need to know how to write expressions of the form $a \cos x \pm b \sin x$ in the form $R \cos (x \pm \alpha)$ or $R \sin (x \pm \beta)$ . You should also be able to solve equations of the form $a \cos x \pm b \sin x = c$ . The addition formulae are used to help derive these expressions.

$R \cos (x \pm \alpha)$ and $R \sin(x \pm \beta)$

Use the addition formulae to write expressions of the form $a \cos x \pm b \sin x$ in the form $R \cos (x \pm \alpha)$ or $R \sin(x \pm \beta)$ . Once the addition formulae has been applied, sketch a right-angled triangle to find $R$ and $\alpha$ or $\beta$ .

Example 1

Write the expression $4 \sin x + 2 \cos x$ in the form $R \sin (x+ \alpha)$ .

Use the addition formulae on $R \sin (x+ \alpha)$ :

$\begin{aligned}4 \sin x + 2 \cos x &= R \sin (x+\alpha) \\4 \sin x + 2 \cos x &= R [\sin x \cos \alpha+ \cos x \sin \alpha] \\4 \sin x + 2 \cos x &= R \sin x \cos \alpha+ R\cos x \sin \alpha \\\end{aligned}$

Equate the coefficients of $\sin x$ and $\cos x$ . Then draw a sketch to find $R$ and $\alpha$ using trigonometric rules:

$4 = R \cos \alpha \\2 = R \sin \alpha$
$R = \sqrt{4^2+2^2}= 2\sqrt5$	$\begin{aligned}\tan \alpha &= \dfrac{2}{4}\\ \\\alpha& = 26.6\degree \end{aligned}$

Therefore, $\underline{4 \sin x + 2 \cos x = 2 \sqrt5 \sin (x+26.6\degree)}$ .

Solving equations with $a\cos x + b \sin x$

You should be able to solve equations of the form $a\cos x + b \sin x=c$ . First use the above method to write the equation in the form $R \sin (x+\alpha)$ or $R \cos (x+\alpha)$ before solving the trigonometric equation. A sketch of the graph may help.

Example 2

The expression $4 \cos x + 7 \sin x$ can be written in the form $R \cos (x-\alpha)$ .

a) Find $R$ and $\alpha$ .

b) Find the maximum value of the expression $R \cos (x-\alpha)$ and the minimum value of $x$ for which this value occurs in the range $0 \le x \le 360\degree$ .

c) Solve $4 \cos x + 7 \sin x=2$ in the range $0 \le x \le 360\degree$ .

a) Use the addition formula on $R \cos (x-\alpha)$ :

$\begin{aligned}4 \cos x + 7 \sin x &= R \cos (x-\alpha) \\4 \cos x + 7 \sin x &= R [\cos x \cos \alpha + \sin x \sin \alpha] \\4 \cos x + 7 \sin x &= R \cos x \cos \alpha + R\sin x \sin \alpha \\\end{aligned}$

Equate coefficients to find $R$ and $\alpha$ :

$\begin{aligned}4 &= R \cos \alpha \\7 &= R \sin \alpha \\\\R &= \sqrt{4^2+7^2} \\R &= \underline{\sqrt{65}}\\\\\tan \alpha&= \dfrac 7 4 \\ \\\alpha&= \underline{60.3 \degree} \ (3 \ s.f.)\end{aligned}$

b) The expression is therefore $\sqrt{65} \cos (x-60.3\degree)$ and the maximum value occurs when $\cos(x-60.3\degree) = 1$ , so is $\sqrt{65} \times 1$ .

Therefore, the maximum value of the expression is $\underline{\sqrt{65}}$ .

To find the minimum value of $x$ for which this occurs, equate the expression to $\sqrt{65}$ and solve:

$\begin{aligned}\sqrt{65} \cos (x-60.3\degree) &= \sqrt{65} \\\cos (x-60.3\degree) &= 1 \\x-60.3\degree &= \cos^{-1} (1) \\x-60.3\degree &= 0 \\x&= \underline{60.3\degree}\end{aligned}$ 

c) Substitute the expression $\sqrt{65} \cos (x-60.3\degree)$ in and solve:

$\begin{aligned}4 \cos x + 7 \sin x&=2 \\\sqrt{65} \cos (x-60.3\degree)& = 2 \\\cos (x-60.3\degree) &= \dfrac{2}{\sqrt{65}} \\x-60.3\degree &= \cos^{-1} \Big(\dfrac{2}{\sqrt{65}}\Big) \\x-60.3\degree &= 75.64 , 284.36 \\x &= \underline{135.9 \degree, 344.7 \degree \ (1 \ d.p.)}\end{aligned}$