Question

2) evaluate each of the following in each case, and explain
a. cos^-1(cos(-pi/6)
part1: what is the value of cos(-pi/6)
part2; to solve the problem cos^-1(cos(-pi/6))find the angle in interval [0,pi] whose cosine is sqrt3/2.

b. sin(tan^-1(4/3))

part1: find the measure of the hypotenuse of this triangle.
part2: angle (theta) is the angle whose tan is 4/3. find the sine of angle (theta)

c. tan^-1(cos(pi/2))

part1: what is the value of cos(pi/2)
part2: to solve tan^-1(cos(pi/2)) find the angle in the interval [-pi/2, pi/2] whose tangent equals 0.

Answer 1

\[\cos (-\frac{\pi}{6})=\frac{\sqrt{3}}{2}\]

Answer 2

The desired angle is
\[\frac{\pi}{6}\]