# Precalculus : Evaluate Expressions That Include the Inverse Tangent, Cosecant, Secant, or Cotangent Function

## Example Questions

### Example Question #31 : Graphs And Inverses Of Trigonometric Functions

Approximate:

Explanation:

:

There is a restriction for the range of the inverse tangent function from .

The inverse tangent of a value asks for the angle where the coordinate  lies on the unit circle under the condition that .  For this to be valid on the unit circle, the  must be very close to 1, with an  value also very close to zero, but cannot equal to zero since  would be undefined.

The point  is located on the unit circle when  , but  is invalid due to the existent asymptote at this angle.

An example of a point very close to  that will yield  can be written as:

Therefore, the approximated rounded value of  is .

### Example Question #1 : Inverse Trigonometric Functions

Evaluate:

Explanation:

To determine the value of , solve each of the terms first.

The inverse cosine has a domain and range restriction.

The domain exists from , and the range from .  The inverse cosine asks for the angle when the x-value of the existing coordinate is .  The only possibility is  since the coordinate can only exist in the first quadrant.

The inverse sine also has a domain and range restriction.

The domain exists from , and the range from .  The inverse sine asks for the angle when the y-value of the existing coordinate is .  The only possibility is  since the coordinate can only exist in the first quadrant.

Therefore:

### Example Question #1 : Inverse Trigonometric Functions

Determine the correct value of  in degrees.

Explanation:

Rewrite and evaluate .

The inverse sine of one-half is  since  is the y-value of the coordinate when the angle is .

To convert from radians to degrees, replace  with 180.

### Example Question #3 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

Evaluate the following:

Explanation:

For this particular problem we need to recall that the inverse cosine cancels out the cosine therefore,

.

So the expression just becomes

From here, recall the unit circle for specific angles such as .

Thus,

.