If a and b are positive, asinθ±bcosθ can be written in the form Rsin(θ±α), acosθ±bsinθ can be written in the form Rsin(θ∓α), where R=√a2+b2,Rcosα=a,Rsinα=b and tanα=ba with 0∘<α<90∘. |
Example (1) Solve the equation 8sinθ+6cosθ=5 for 0∘≤θ≤360∘.
Solution
8sinθ+6cosθ=5
Let Rcosα=8 and Rsinα=6.
∴ 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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