Trigonometry! #16

Let $x = \theta + \frac{\pi}{6}.$ Then the expression becomes

$S = 5\cos(x - \frac{\pi}{6}) + 3\cos(x + \frac{\pi}{6}) + 3 =$

$5(\cos(x)\cos(\frac{\pi}{6}) + \sin(x)\sin(\frac{\pi}{6})) + 3(\cos(x)\cos(\frac{\pi}{6}) - \sin(x)\sin(\frac{\pi}{6})) + 3 =$

$\frac{5}{2}(\sqrt{3}\cos(x) + \sin(x)) + \frac{3}{2}(\sqrt{3}\cos(x) - \sin(x)) + 3 =$

$4\sqrt{3}\cos(x) + \sin(x) + 3.$

Now note that $\sqrt{(4\sqrt{3})^{2} + 1^{2}} = \sqrt{48 + 1} = 7.$ So if we let $\alpha = \cos^{-1}(\frac{1}{7})$ then $\cos(\alpha) = \frac{1}{7}$ and $\sin(\alpha) = \frac{4\sqrt{3}}{7}.$ This allows us to write $S$ as

$7(\sin(\alpha)\cos(x) + \cos(\alpha)\sin(x)) + 3 = 7\sin(\alpha + x) + 3.$

Now since $(\alpha + x)$ can be any value, we have that $-1 \le \sin(\alpha + x) \le 1$ , so

$(-7 + 3) \le S \le (7 + 3) \Longrightarrow -4 \le S \le 10.$

The sum of the minimum and maximum values is thus $-4 + 10 = \boxed{6}$ .

First of all, in order to know the angles which will give us the minimum and maximum values, we have to get the first derivative of the expression with respect to theta, which is: $-5\sin \theta - 3\sin\left(\theta + \frac {\pi}{3}\right)$ Now by substituting the value of the special angles, we can find the angles of the minimum ad maximum values, which are angles 90 and 270. By substituting these angles in the original expression we get that: $5\cos\ 90 +3\cos\ 150 +3$ = $\frac{\ 6-3 \sqrt{3}}{2}$ and $5\cos\ 270 +3\cos\ 330 +3$ = $\frac{\ 6+3 \sqrt{3}}{2}$ By adding both values, we'll get that the sum of the minimum and maximum values is $\boxed{6}$