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

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Question

Find angle A of the following triangle:

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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:

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?

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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($\frac{\sqrt3}{2}\right)$

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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)$

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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)$ .

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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 .

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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:

Tap to reveal answer

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:

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?

Tap to reveal answer

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($\frac{\sqrt3}{2}\right)$

Tap to reveal answer

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)$

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Find the inverse of the function

.

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

Tap to reveal answer

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}$)=y$

← Didn't Know|Knew It →

Question

Find angle A of the following triangle:

Tap to reveal answer

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:

which becomes

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

← Didn't Know|Knew It →

Question

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 reveal answer

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($\frac{\sqrt3}{2}\right)$

Tap to reveal answer

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)$

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Find the inverse of the function

.

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

Tap to reveal answer

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}$)=y$

← Didn't Know|Knew It →

Question

Find angle A of the following triangle:

Tap to reveal answer

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:

which becomes

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

← Didn't Know|Knew It →

Question

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 reveal answer

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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