Evaluate Expressions That Include the Inverse Sine or Cosine Function - Pre-Calculus

Card 0 of 24

Question

Find angle A of the following triangle:

Answer

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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Question

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

Answer

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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Question

Evaluate:

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

Answer

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 domain is only valid from .

$\theta$ 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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Question

Evaluate:

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

Answer

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:

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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Question

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

Answer

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:

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 $x=\frac{1}{2}$ is 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 $\frac{1}{2}$ is degrees.

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Question

Find the inverse of the function

.

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

Answer

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 exchange and ,

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

Solving for

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

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

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Question

Find angle A of the following triangle:

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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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Question

Evaluate:

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

Answer

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 domain is only valid from .

$\theta$ 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:

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

Compare your answer with the correct one above

Question

Evaluate:

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

Answer

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:

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 $\frac{\pi}{3}$ .

Compare your answer with the correct one above

Question

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

Answer

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:

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 $x=\frac{1}{2}$ is 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 $\frac{1}{2}$ is degrees.

Compare your answer with the correct one above

Question

Find the inverse of the function

.

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

Answer

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 exchange and ,

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

Solving for

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

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

Compare your answer with the correct one above

Question

Find angle A of the following triangle:

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

Evaluate:

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

Answer

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 domain is only valid from .

$\theta$ 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:

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

Compare your answer with the correct one above

Question

Evaluate:

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

Answer

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:

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 $\frac{\pi}{3}$ .

Compare your answer with the correct one above

Question

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

Answer

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:

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 $x=\frac{1}{2}$ is 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 $\frac{1}{2}$ is degrees.

Compare your answer with the correct one above

Question

Find the inverse of the function

.

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

Answer

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 exchange and ,

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

Solving for

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

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

Compare your answer with the correct one above

Question

Find angle A of the following triangle:

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

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