Precalculus : Inverse Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

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Example Questions

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Example Question #1 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

Approximate:  

Possible Answers:

Correct answer:

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:  

Possible Answers:

Correct answer:

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

Evaluate the following:

Possible Answers:

Correct answer:

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,

.

Example Question #1 : Inverse Trigonometric Functions

Determine the value of  in degrees.

Possible Answers:

Correct answer:

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 #851 : Pre Calculus

Evaluate the following expression: 

Possible Answers:

Correct answer:

Explanation:

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

 

Example Question #1 : Inverse Trigonometric Functions

Evaluate: 

Possible Answers:

Correct answer:

Explanation:

 

Example Question #31 : Graphs And Inverses Of Trigonometric Functions

Approximate the following: 

Possible Answers:

Correct answer:

Explanation:

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

 

Example Question #1 : Inverse Trigonometric Functions

Given that   and that  is acute, find the value of  without using a calculator.

Possible Answers:

Correct answer:

Explanation:

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) 

 

Example Question #1 : Inverse Trigonometric Functions

Given that  and that  is acute, find the value of  without using a calculator.

Possible Answers:

Correct answer:

Explanation:

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 .

Example Question #1 : Inverse Trigonometric Functions

Evaluate the following expression: 

Possible Answers:

Correct answer:

Explanation:

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 .

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