Trigonometrical Equations 3 Question 20
21. Prove that $5 \cos \theta+3 \cos \theta+\frac{\pi}{3}+3$ lies between -4 and 10.
(1979, 3M)
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Solution:
- Let $f(\theta)=5 \cos \theta+3 \cos \theta+\frac{\pi}{3}+3$
$ =5 \cos \theta+3 \cos \theta \cos \frac{\pi}{3}-\sin \theta \sin \frac{\pi}{3}+3 $
$ \begin{aligned} & =5 \cos \theta+3 \frac{1}{2} \quad \cos \theta-3 \frac{\sqrt{3}}{2} \sin \theta+3 \\ & =\frac{13}{2} \cos \theta-\frac{3 \sqrt{3}}{2} \sin \theta+3 \\ \Rightarrow \quad f(\theta) & =\frac{1}{2}(13 \cos \theta-3 \sqrt{3} \sin \theta)+3 \end{aligned} $
Put $r \cos \alpha=13, r \sin \alpha=3 \sqrt{3}$, then
$ \begin{aligned} r & =\sqrt{169+27} \\ & =\sqrt{196}=14 \\ \therefore \quad f(\theta) & =\frac{1}{2}(r \cos \alpha \cos \theta-r \sin \alpha \sin \theta)+3 \\ & =\frac{1}{2} r \cos (\theta+\alpha)+3 \\ & =7 \cos (\theta+\alpha)+3 \end{aligned} $
Now, $\quad-1 \leq \cos (\theta+\alpha) \leq 1$
$\Rightarrow \quad-7 \leq 7 \cos (\theta+\alpha) \leq 7$
$\Rightarrow \quad-4 \leq f(\theta) \leq 10$
