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Precalculus : Inverse Sine and Cosine Functions

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Example Questions

Example Question #1 :Inverse Sine And Cosine Functions

Find angle A of the following triangle:

Using_inverse_sin_to_find_angle_of_triangle

Possible Answers:

$57^{\circ}$

$13^\circ$

$29^{\circ}$

None of the other answers

$33^{\circ}$

Correct answer:

$33^{\circ}$

Explanation:

We are given the hypotenuse and the side opposite of the angle in question. The trig function that relates these two sides is SIN. Therefore, we can write:

$sin(A) = \frac{12}{22}$

In order to solve for A, we need to take the inverse sin of both sides:

$sin^{-1}(sin(A)) = sin^{-1}(\frac{12}{22})$

which becomes

$A = sin^{-1}\bigg(\frac{12}{22}\bigg) = 33^{\circ}$

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Example Question #1 :Inverse Sine And Cosine Functions

Consider $sin(\theta)=-\frac{1}{2}$ , where theta is valid from $\left [ 0,2\pi\right ]$ . What is a possible value of theta?

Possible Answers:

$-\frac{\pi}{6}$

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

$\frac{11\pi}{6}$

$None\: of \: the\: other\: answers.$

$-\frac{\pi}{3}$

Correct answer:

$\frac{11\pi}{6}$

Explanation:

Solve for theta by taking the inverse sine of both sides.

$\theta=sin^-^1\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$

Since this angle is not valid for the given interval of theta, add $2\pi$ radians to this angle to get a valid answer in the interval.

$-\frac{\pi}{6}+2\pi = -\frac{\pi}{6}+\frac{12\pi}{6}= \frac{11\pi}{6}$

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Example Question #3 :Inverse Sine And Cosine Functions

Evaluate:

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

Possible Answers:

$-\pi$

$\pi$

Correct answer:

$\pi$

Explanation:

First evaluate $cos^-^1\left(-\frac{\sqrt3}{2}\right)$ .

To evaluate inverse cosine, it is necessary to know the domain and range of inverse cosine.

For: $\theta=cos^-^1(x)$

The domainis only valid from [-1,1] .

$\theta$ is only valid from $[0,\pi]$ .

The part is asking for the angle where the x-value of the coordinate is $-\frac{\sqrt3}{2}$ . The only possibility on the unit circle is the second quadrant.

$cos^-^1\left(-\frac{\sqrt3}{2}\right)= \frac{5}{6}\pi$

Next, evaluate $cos^-^1\left(\frac{\sqrt3}{2}\right)$ .

Using the same domain and range restrictions, the only valid angle for the given x-value is in the first quadrant on the unit circle.

$cos^-^1\left(\frac{\sqrt3}{2}\right)= \frac{\pi}{6}$

Therefore:

$cos^-^1\left(-\frac{\sqrt3}{2}\right)+cos^-^1\left(\frac{\sqrt3}{2}\right) = \frac{5}{6}\pi + \frac{\pi}{6}=\frac{6\pi}{6}=\pi$

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Example Question #1 :Inverse Sine And Cosine Functions

Evaluate:

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

Possible Answers:

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

$\frac{\pi}{5}$

$\frac{\pi}{3}$

$\pi$

$\frac{\pi}{6}$

Correct answer:

$\frac{\pi}{3}$

Explanation:

To find the correct value of $cos^-^1\left(\frac{1}{2}\right)$ , it is necessary to know the domain and range of inverse cosine.

Domain: [-1,1]

Range: $[0,\pi]$

The question is asking for the specific angle when the x-coordinate is half.

The only possibility is located in the first quadrant, and the point of the special angle is $\left(\frac{1}{2},\frac{\sqrt3}{2}\right)$

The special angle for this coordinate is $\frac{\pi}{3}$ .

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Example Question #5 :Inverse Sine And Cosine Functions

Find the value of $\cos^-^1\left(\frac{1}{2}\right)$ .

Possible Answers:

$-60^\circ$

$300^\circ$

$30^\circ$

$60^\circ$

$120^\circ$

Correct answer:

$60^\circ$

Explanation:

In order to determine the value or values of $cos^-^1(\frac{1}{2})$ , it is necessary to know the domain and range of the inverse sine function.

Domain: [-1,1]

Range: $[0,\pi]$

The question is asking for the angle value of theta where the x-value is $\frac{1}{2}$ under the range restriction. Since $x=\frac{1}{2}$ 是located in the first and fourth quadrants, the range restriction makes theta only allowable from $[0,\pi]$ . Therefore, the theta value must only be in the first quadrant.

The value of the angle when the x-value is $\frac{1}{2}$ isdegrees.

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Example Question #6 :Inverse Sine And Cosine Functions

Find the inverse of the function

y=csc(x)tan(x) .

Make sure the final notation is only in the forms including arcsin , arccos , and arctan .

Possible Answers:

$arccos(\frac{1}{x})=y$

$y=arctan(arcsin(\frac{1}{x}))$

$y=arcsin(\frac{1}{x})$

arcsec(x)=y

Correct answer:

$arccos(\frac{1}{x})=y$

Explanation:

The easiest way to solve this problem is to simplify the original expression.

$y=csc(x)*tan(x)=\frac{1}{sin(x)}*\frac{sin(x)}{cos(x)}= \frac{1}{cos(x)}$

$y=\frac{1}{cos(x)}$

To find its inverse, let's exchangeand,

$x=\frac{1}{cos(y)}$

Solving for

$\frac{1}{x}=cos(y)$

$arccos(\frac{1}{x})=y$

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Precalculus : Inverse Sine and Cosine Functions

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