| 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}'$.
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