Evaluate Expressions That Include the Inverse Tangent, Cosecant, Secant, or Cotangent Function - Pre-Calculus

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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 .

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.

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,

.

This one seems complicated, but becomes considerably easier once you implement the fact that the composite cancels out to 1 and you are left with which is equal to 1

This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which always

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of (the adjacent side over the opposite side)

This one seems complicated but becomes considerably easier once you implement the fact that the composite cancels out to and you are left with which is equal to , and so the answer is .

and so the credited answer is .

This problem is quite manageable with knowledge of the unit circle: the value is extremely close to zero, of which always, so the only reasonable estimation of this value is 0.

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): .

From here we can deduce the value of (the adjacent side over the opposite side) and so the answer is .

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:

:

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 .

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.

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,

.

This one seems complicated, but becomes considerably easier once you implement the fact that the composite cancels out to 1 and you are left with which is equal to 1

This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which always

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of (the adjacent side over the opposite side)