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:

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