Home > Trigonometry > Basic Inverse Trigonometric Functions

How do you evaluate #Arccos (- sqrt 3 / 2)#?

Answer 1

#(5pi)/6 and (7pi)/6#

Trig table and unit circle --> #cos x = (-sqrt3/2)# --> arc #x = +- (5pi)/6# Answers for # (0, 2pi)#; #(5pi)/6 and (7pi)/6# (co-terminal to # -(5pi)/6#)

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Madelyn Adams

To evaluate arccos(-√3/2), first note that the arccosine function returns an angle whose cosine is the given value. Since the cosine function is negative in the second and third quadrants, we're looking for an angle in either of those quadrants where the cosine is -√3/2. The cosine of π/6 radians (or 30 degrees) is √3/2, but since we're looking for a negative value, we take the angle in the second quadrant, so arccos(-√3/2) = 2π/3 radians (or 120 degrees).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.