NSW Y12 Maths - Extension 1 Trig Functions Auxiliary Angle Method

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Auxiliary Angle Method Theory

$Writing the equation $a \sin x+b \cos x=c$ in the form $r \sin (x+\alpha)=c $ or $ r\cos (x-\alpha)=c$ allows you to solve the equation. \\ \textbf{Example}\\ In the domain, $0 \le \theta \le 2 \pi$, solve $\sqrt{3} \cos \theta - \sin \theta =1$\\ \textbf{Solution}\\ $\begin{aligned} \text{Let } \sqrt{3} \cos \theta-\sin \theta & \equiv A \cos (\theta+\alpha)=1 \\ & \equiv A \cos \theta \cos \alpha-A \sin \theta \sin \alpha \end{aligned}$\\ $\begin{aligned} A \cos \alpha &=\sqrt{3} & A \sin \alpha &=1 \\ \cos \alpha &=\frac{\sqrt{3}}{A} & \sin \alpha &=\frac{1}{A} \\ \tan \alpha &=\frac{1}{\sqrt{3}} & \frac{3}{A^2} &+\frac{1}{A^2}=1 \\ \alpha &=\frac{\pi}{6} & A^2 &=4 \\ & &A =2 \end{aligned}$\\ $\begin{aligned} \therefore \quad 2 \cos \left(\theta+\frac{\pi}{6}\right)=1 \\ \quad \cos \left(\theta+\frac{\pi}{6}\right)&=\frac{1}{2} \\ \theta+\frac{\pi}{6}&=\frac{\pi}{3}, \frac{5 \pi}{3} \\ \therefore \theta&=\frac{\pi}{6}, \frac{3 \pi}{2} \end{aligned}$\\$

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