# Equation of the Type : a sin θ ± b cos θ = c

 If a and b are positive,                          $\displaystyle a\sin \theta \pm b\cos \theta$ can be written in the form $\displaystyle R\sin (\theta \pm \alpha ),$                          $\displaystyle a\cos \theta \pm b\sin \theta$ can be written in the form $\displaystyle R\sin (\theta \mp \alpha ),$  where $R = \sqrt{a^2 + b^2}, R \cos \alpha = a , R \sin \alpha = b$ and $\displaystyle \tan \alpha =\frac{b}{a}$ with $\displaystyle {{0}^{{}^\circ }}<\alpha <{{90}^{{}^\circ }}.$

Example (1)      Solve the equation $\displaystyle 8\sin \theta +6\cos \theta =5$ for $\displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$.

Solution

$\displaystyle 8\sin \theta +6\cos \theta =5$

Let $\displaystyle R\cos \alpha =8$ and $\displaystyle R\sin \alpha =6$.

$\displaystyle \therefore R=\sqrt{{{{8}^{2}}+{{6}^{2}}}}=\sqrt{{100}}=10$ and $\displaystyle \tan \alpha =\frac{6}{8}\Rightarrow \alpha =36{}^\circ 5{2}'$

Since $\displaystyle 8\sin \theta +6\cos \theta =R\sin (\theta +\alpha )$,

$\displaystyle R\sin (\theta +\alpha )=5\Rightarrow 10\sin (\theta +36{}^\circ 5{2}')=5\Rightarrow \sin (\theta +36{}^\circ 5{2}')=\frac{1}{2}$

$\displaystyle \begin{array}{l}\therefore \theta +36{}^\circ 5{2}'=30{}^\circ \\\end{array}$ (1st quadrant) or

$\displaystyle \theta +36{}^\circ 5{2}'=150{}^\circ$(2nd quadrant) or

$\displaystyle \theta +36{}^\circ 5{2}'=390{}^\circ$(1st quadrant)

$\displaystyle \therefore \theta =-6{}^\circ 5{2}'$ or $\displaystyle \ \theta =113{}^\circ {8}'$ or $\displaystyle \theta =353{}^\circ {8}'$

Since $\displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$, $\displaystyle \theta =-6{}^\circ 5{2}'\$is impossible.

$\displaystyle \therefore \theta =113{}^\circ {8}'$ or $\displaystyle \theta =353{}^\circ {8}'$.