Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 23

cot(arcsin (-1/2))
To evaluate this, let's first consider the innermost expression. Let it be equal to y.
y = arcsin(-1/2)
Rewriting it in terms of sine function, the equation becomes
sin (y) = -1/2
Base on Table of Special Angles (see attached), sine has a value of -1/2 at angles (7pi)/6 and (11pi)/6 . So the values of angle y are
y = (7pi)/6, (11pi)/6
Then, let's refer to the arcsine again.
y = arcsin(-1/2)
Take note that the range of arcsine is -pi/2lt= y lt=pi/2 . Neither (7pi)/6 nor (11pi)/6 belong to this interval. So let's consider the negative angle that is coterminal of these two.
To determine the negative angles that are coterminal with (7pi)/6 and (11pi)/6 , subtract 2pi from each angle.
(7pi)/6 - 2pi = -(5pi)/6
(11pi)/6-2pi=-pi/6
Between these two values, it is only -pi/6 that belongs to the interval -pi/2lt= ylt=pi/2 .So the value of angle y is
y = arcsin (-1/2) = -pi/6
Substituting this to the original expression
cot(arcsin (-1/2))
it becomes
= cot (-pi/6)
Applying the negative angle identity cot (-theta)=-cot (theta) , it simplifies to
=-cot(pi/6)
=-sqrt3

Therefore, cot (arcsin (-1/2)) = -sqrt3 .