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

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Find angle A of the following triangle:

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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:

which becomes

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

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

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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}$$

Evaluate:

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

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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}$=\pi$

Evaluate:

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

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

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

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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.

Find the inverse of the function

.

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

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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}$)=y$

Find angle A of the following triangle:

Tap to see back →

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:

which becomes

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

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

Tap to see back →

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}$$

Evaluate:

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

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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}$=\pi$

Evaluate:

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

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

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

Tap to see back →

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.

Find the inverse of the function

.

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

Tap to see back →

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}$)=y$

Find angle A of the following triangle:

Tap to see back →

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:

which becomes

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

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

Tap to see back →

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}$$

Evaluate:

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

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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}$=\pi$

Evaluate:

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

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

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

Tap to see back →

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.

Find the inverse of the function

.

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

Tap to see back →

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}$)=y$

Find angle A of the following triangle:

Tap to see back →

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:

which becomes

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

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

Tap to see back →

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}$$