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Precalculus : Inverse Trigonometric Functions

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Example Questions

Example Question #1 : Inverse Trigonometric Functions

Approximate: tan^-^1(99999)

Possible Answers:

$\infty$

$\frac{\pi}{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{\pi}{2}$

Explanation:

tan^-^1(99999) :

There is a restriction for the range of the inverse tangent function from $(-\frac{\pi}{2},\frac{\pi}{2})$ .

The inverse tangent of a value asks for the angle where the coordinate (x,y) lies on the unit circle under the condition that $\frac{y}{x}=99999$ . 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 $\frac{y}{x}$ would be undefined.

The point (0,1) is located on the unit circle when $\theta=\frac{\pi}{2}$ , but $tan(\frac{\pi}{2})$ is invalid due to the existent asymptote at this angle.

An example of a point very close to (0,1) that will yield $\frac{y}{x}=99999$ can be written as:

(0.00001,0.99999)

Therefore, the approximated rounded value of tan^-^1(99999) is $\frac{\pi}{2}$ .

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Example Question #2 : Inverse Trigonometric Functions

Evaluate:

$cos^-^1\left(\frac{\sqrt3}{2}\right)+sin^-^1\left(\frac{\sqrt3}{2}\right)=$

Possible Answers:

$\frac{\pi}{3}$

$\frac{\pi}{2}$

$\frac{3\pi}{2}$

Correct answer:

$\frac{\pi}{2}$

Explanation:

To determine the value of $cos^-^1(\frac{\sqrt3}{2})+sin^-^1(\frac{\sqrt3}{2})$ , solve each of the terms first.

$cos^-^1(\frac{\sqrt3}{2})$

The inverse cosine has a domain and range restriction.

The domain exists from [-1,1] , and the range from $[0,\pi]$ . The inverse cosine asks for the angle when the x-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{6}$ since the coordinate can only exist in the first quadrant.

$sin^-^1(\frac{\sqrt3}{2})$

The inverse sine also has a domain and range restriction.

The domain exists from [-1,1] , and the range from $[-\frac{\pi}{2},\frac{\pi}{2}]$ . The inverse sine asks for the angle when the y-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{3}$ since the coordinate can only exist in the first quadrant.

Therefore:

$cos^-^1(\frac{\sqrt3}{2})+sin^-^1(\frac{\sqrt3}{2})= \frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$

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Example Question #1 : Inverse Trigonometric Functions

Evaluate the following:

$f(x)=cos(cos^{-1}(cos(\frac{\pi }{3})))$

Possible Answers:

2.537

$\frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}2{}$

1.34235

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

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

$cos^{-1}(cos(\theta ))=\theta$ .

So the expression just becomes

$f(x)=cos\left(\frac{\pi }{3}\right)$

From here, recall the unit circle for specific angles such as $\frac{\pi}{3}$ .

Thus,

$f(x)=cos\left(\frac{\pi }{3}\right)=\frac{1}{2}$ .

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Example Question #2 : Inverse Trigonometric Functions

Determine the value of 2csc^-^1(2) in degrees.

Possible Answers:

$15^{\circ}$

$90^{\circ}$

$30^{\circ}$

$\textup{The answer is not given.}$

$60^{\circ}$

Correct answer:

$60^{\circ}$

Explanation:

Rewrite and evaluate csc^-^1(2) .

$csc^-^1(2)=sin^-^1(\frac{1}{2})=\frac{\pi}{6}$

The inverse sine of one-half is $\frac{\pi}{6}$ since $\frac{1}{2}$ is the y-value of the coordinate when the angle is $\frac{\pi}{6}$ .

$2csc^-^1(2)=2(\frac{\pi}{6})=\frac{\pi}{3}$

To convert from radians to degrees, replace $\pi$ with 180.

$\frac{\pi}{3}=\frac{180}{3}=60$

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Example Question #4 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

Evaluate the following expression: $f(x)=tan(arctan(tan(\frac{\pi}{4})))$

Possible Answers:

$\frac{\sqrt{2}}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

This one seems complicated, but becomes considerably easier once you implement the fact that the composite tan(arctan) cancels out to 1 and you are left with $tan(\frac{\pi}{4})$ which is equal to 1

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

Evaluate: $5sin^{-1}(\frac{\sqrt{3}}{2})$

Possible Answers:

$\frac{5\pi}{2}$

$\frac{5\sqrt{3}}{2}$

$\frac{\pi}{3}$

$\pi$

$\frac{5\pi}{3}$

Correct answer:

$\frac{5\pi}{3}$

Explanation:

$arcsin(\frac{\sqrt{3}}{2})=60^\circ\rightarrow \frac{\pi}{3}\rightarrow 5\cdot \frac{\pi}{3}$

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Example Question #2 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

Approximate the following: arcsin(0.00002)

Possible Answers:

0.2

Correct answer:

Explanation:

This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which arcsin(0)=0 always

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Example Question #854 : Pre Calculus

Given that $sin\theta=\frac{3}{4}$ and that $\theta$ is acute, find the value of $cot\theta$ without using a calculator.

Possible Answers:

$\frac{\sqrt{7}}{3}$

$4\sqrt{7}$

$\sqrt{7}$

$\frac{3}{\sqrt{7}}$

$3\sqrt{7}$

Correct answer:

$\frac{\sqrt{7}}{3}$

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”): $t=\sqrt{4^{2}-3^{2}}\rightarrow t=\sqrt{7}$ . From here we can easily deduce the value of $cot\theta=\frac{\sqrt{7}}{3}$ (the adjacent side over the opposite side)

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

Given that $\sin \theta =\frac{3}{4}$ and that $\theta$ is acute, find the value of $\cot \theta$ without using a calculator.

Possible Answers:

$4\sqrt{7}$

$\sqrt{7}$

$\frac{3}{\sqrt{7}}$

$\frac{\sqrt{7}}3{}$

$3\sqrt{7}$

Correct answer:

$\frac{\sqrt{7}}3{}$

Explanation:

From here we can deduce the value of $\cot \theta =\frac{\sqrt{7}}{3}$ (the adjacent side over the opposite side) and so the answer is $\frac{\sqrt{7}}{3}$ .

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Example Question #5 : Inverse Trigonometric Functions

Evaluate the following expression: $f\left ( x \right )=\tan (\arctan (\tan (\frac{\pi }{4})))$

Possible Answers:

$\frac{\sqrt{2}}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

This one seems complicated but becomes considerably easier once you implement the fact that the composite $\tan \left ( \arctan \right )$ cancels out to and you are left with $\tan \left ( \frac{\pi }{4} \right )$ which is equal to , and so the answer is .

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Precalculus : Inverse Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Inverse Trigonometric Functions

Example Question #2 : Inverse Trigonometric Functions

Example Question #1 : Inverse Trigonometric Functions

Example Question #2 : Inverse Trigonometric Functions

Example Question #4 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

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

Example Question #2 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function

Example Question #854 : Pre Calculus

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

Example Question #5 : Inverse Trigonometric Functions

All Precalculus Resources

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