Question

Prove that $5\cos \theta +3\cos (\theta +\pi /3)+3$ lies between $-4$ and $10$.

Answer 1

The given expression is
$5\cos \theta +3(\cos \theta \cos {60}^o-\sin \theta \sin {60}^o)+3=\frac{1}{2}[13\cos \theta -3\sqrt{3}\sin \theta ]+3$
Put $13=r\cos \alpha ,3\sqrt{3}=r\sin \alpha$
$\therefore r=\sqrt{169+27}=\sqrt{196}=14$
Given expression
$=\frac{r}{2}\left[\cos (\theta +\alpha )\right]+3=7\cos (\theta +\alpha )+3$
Hence max. and min. value of expression are $7+3$ and $-7+3$ i.e., $10$ and $-4$ respectively