Question
Let $$p = a\cos \theta - b\sin \theta .$$ Then for all real $$\theta $$
A.
$$p > \sqrt {{a^2} + {b^2}} $$
B.
$$p < - \sqrt {{a^2} + {b^2}} $$
C.
$$ - \sqrt {{a^2} + {b^2}} \leqslant p \leqslant \sqrt {{a^2} + {b^2}} $$
D.
None of these
Answer :
$$ - \sqrt {{a^2} + {b^2}} \leqslant p \leqslant \sqrt {{a^2} + {b^2}} $$
Solution :
$$\eqalign{
& p = \sqrt {{a^2} + {b^2}} \left\{ {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\cos \theta - \frac{b}{{\sqrt {{a^2} + {b^2}} }}\sin \theta } \right\} \cr
& p = \sqrt {{a^2} + {b^2}} \cos \left( {\theta + \alpha } \right),\,\,{\text{where }}\cos \alpha = \frac{a}{{\sqrt {{a^2} + {b^2}} }}. \cr
& {\text{But }} - 1 \leqslant \cos \left( {\theta + \alpha } \right) \leqslant 1. \cr} $$