Linear equation by the added-angle method

Statement

• (a) Express $6\cos\theta+8\sin\theta$ in the form $R\cos(\theta-\alpha)$, with $R>0$ and $0^\circ<\alpha<90^\circ$.
• (b) Solve $6\cos\theta+8\sin\theta=5$ for $0^\circ\le\theta\le 360^\circ$.

Solution

(a) $R=\sqrt{6^2+8^2}=10$ and $\tan\alpha=\dfrac{8}{6}=\dfrac{4}{3}\Rightarrow \alpha\approx 53{.}13^\circ$. Hence $6\cos\theta+8\sin\theta=10\cos(\theta-53{.}13^\circ)$.

(b) $10\cos(\theta-53{.}13^\circ)=5\Rightarrow \cos(\theta-53{.}13^\circ)=\dfrac12\Rightarrow \theta-53{.}13^\circ=\pm 60^\circ$.

So $\theta=113{.}13^\circ$ or $\theta=-6{.}87^\circ\equiv 353{.}13^\circ$.

$\boxed{R=10,\ \alpha\approx 53{.}13^\circ;\quad \theta\approx 113{.}13^\circ,\ 353{.}13^\circ}$