# Question #8dc79

##### 1 Answer
Sep 14, 2016

Let
$5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 3 = y$

$\implies 5 \cos x + 3 \cos x \cos \left(\frac{\pi}{3}\right) - 3 \sin x \sin \left(\frac{\pi}{3}\right) + 3 = y$

$\implies 5 \cos x + \frac{3}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x = y - 3$

$\implies \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x = y - 3$

Now dividing both sides by $\sqrt{{\left(\frac{13}{2}\right)}^{2} + {\left(\frac{3 \sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{169}{4} + \frac{27}{4}}$
$= \sqrt{\frac{169 + 27}{4}} = 7$ we get

$\frac{13}{14} \cos x - \frac{3 \sqrt{3}}{14} \sin x = \frac{y - 3}{7}$

Now if $\frac{13}{14} = \cos A \text{ then } \frac{3 \sqrt{3}}{14} = \sin A$

So

$\cos A \cos x - \sin A \sin x = \frac{y - 3}{7}$

$\implies \cos \left(x + A\right) = \frac{y - 3}{7}$

Now as $- 1 \le \cos \left(x + A\right) \le + 1$ we can write

$- 1 \le \frac{y - 3}{7} \le + 1$

$\implies - 7 \le \left(y - 3\right) \le + 7$

$\implies - 7 + 3 \le \left(y - 3 + 3\right) \le + 7 + 3$

$\implies - 4 \le y \le + 10$

So the given expression
lies between -4 and 10.

Proved