The problem asks us to evaluate several trigonometric expressions: a. $sin(390^\circ)$ b. $cos(\frac{10\pi}{3})$ c. $tan(-420^\circ)$ d. $sin(-660^\circ)$ e. $cos(\frac{41\pi}{4})$ f. $tan(\frac{-19\pi}{3})$ g. $4cos(135^\circ)$ h. $\frac{5}{3}cos(300^\circ)$ i. $-2cos(-150^\circ)$

1. Problem Description

The problem asks us to evaluate several trigonometric expressions:

a. $sin(390^\circ)$

b. $cos(\frac{10\pi}{3})$

c. $tan(-420^\circ)$

d. $sin(-660^\circ)$

e. $cos(\frac{41\pi}{4})$

f. $tan(\frac{-19\pi}{3})$

g. $4cos(135^\circ)$

h. $\frac{5}{3}cos(300^\circ)$

i. $-2cos(-150^\circ)$

2. Solution Steps

a. $sin(390^\circ) = sin(390^\circ - 360^\circ) = sin(30^\circ) = \frac{1}{2}$

b. $cos(\frac{10\pi}{3}) = cos(\frac{10\pi}{3} - 2\pi) = cos(\frac{10\pi}{3} - \frac{6\pi}{3}) = cos(\frac{4\pi}{3}) = cos(\pi + \frac{\pi}{3}) = -cos(\frac{\pi}{3}) = -\frac{1}{2}$

c. $tan(-420^\circ) = tan(-420^\circ + 360^\circ) = tan(-60^\circ) = -tan(60^\circ) = -\sqrt{3}$

d. $sin(-660^\circ) = sin(-660^\circ + 2 \cdot 360^\circ) = sin(-660^\circ + 720^\circ) = sin(60^\circ) = \frac{\sqrt{3}}{2}$

e. $cos(\frac{41\pi}{4}) = cos(\frac{41\pi}{4} - 10\pi) = cos(\frac{41\pi - 40\pi}{4}) = cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

f. $tan(\frac{-19\pi}{3}) = tan(\frac{-19\pi}{3} + 6\pi) = tan(\frac{-19\pi + 18\pi}{3}) = tan(\frac{-\pi}{3}) = -tan(\frac{\pi}{3}) = -\sqrt{3}$

g. $4cos(135^\circ) = 4(-\frac{\sqrt{2}}{2}) = -2\sqrt{2}$

h. $\frac{5}{3}cos(300^\circ) = \frac{5}{3}cos(360^\circ - 60^\circ) = \frac{5}{3}cos(60^\circ) = \frac{5}{3}(\frac{1}{2}) = \frac{5}{6}$

i. $-2cos(-150^\circ) = -2cos(-150^\circ + 360^\circ) = -2cos(210^\circ) = -2cos(180^\circ + 30^\circ) = -2(-cos(30^\circ)) = 2cos(30^\circ) = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}$

3. Final Answer

a. $\frac{1}{2}$

b. $-\frac{1}{2}$

c. $-\sqrt{3}$

d. $\frac{\sqrt{3}}{2}$

e. $\frac{\sqrt{2}}{2}$

f. $-\sqrt{3}$

g. $-2\sqrt{2}$

h. $\frac{5}{6}$

i. $\sqrt{3}$