Arcsin, Arccos, Arctan - Trigonometry

Card 0 of 32

What is $\theta$ if and ?

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In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$$

$\tan=\frac{10}{8}$$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow $51.34^{\circ}$$

If you calculated the angle's measure to be then your calculator was set to radians and needs to be set on degrees.

Which of the following is the degree equivalent of the inverse trigonometric function

$\arccos x=\frac{1}{2}$$ ?

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The $\arccos$ is the reversal of the cosine function. That means that if $\cos \theta=y$ , then $\arccos y=\theta$ .

Therefore,

$\theta=\arccos\left($\frac{1}{2}\right)=60^\circ$

For the above triangle, what is $\theta$ if and ?

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We need to use a trigonometric function to find $\theta$ . We are given the opposite and adjacent sides, so we can use the $\tan$ and $\arctan$ functions.

$\tan(\theta) = $\frac{\textup{opposite}$}{\textup{adjacent}}$

$\tan(\theta) = $\frac{20}{14}$$

$\theta = \arctan\left ( $\frac{20}{14}$ \right )$

$\theta = $55.0^{\circ}$$

For the above triangle, what is $\theta$ if and ?

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We need to use a trigonometric function to find $\theta$ . We are given the opposite and hypotenuse sides, so we can use the $\sin$ and $\arcsin$ functions.

$\sin(\theta) = $\frac{\textup{opposite}$}{\textup{hypotenuse}}$

$\sin(\theta) = $\frac{17}{25}$$

$\theta = \arcsin\left ( $\frac{17}{25}$ \right )$

$\theta = $42.8^{\circ}$$

Assuming the angle in degrees, determine the value of .

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To evaluate , it is necessary to know the existing domain and range for these inverse functions.

Inverse sine:

$R \textup{ (in radians)}: [-$\frac{\pi}{2}$,$\frac{\pi}{2}$]$

Inverse cosine:

$R\textup{ (in radians)}: [0,\pi]$

Inverse tangent:

$D: \textup{All real numbers.}$

$R\textup{ (in radians)}: (-$\frac{\pi}{2}$,$\frac{\pi}{2}$)$

Evaluate each term. The final answers must return an angle.

If

$cos(arcsin(sec \ x)) = 0{}$ ,

what value(s) does take?

Assume that $0\leq x< 2\pi$

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If $cos(arcsin(sec \ x)) = 0{}$ , then we can apply the cosine inverse to both sides:

$arcsin(sec \ x)) = $\frac{\pi }{2}$ \ or -$\frac{\pi }{2}$$

Since cosine and cosine inverse undo each other; we can then apply sine and secant inverse functions to obtain the solution.

${}sin(arcsin(sec \ x))) = sin\left($\frac{\pi }{2}$ \right)$ and $sin(arcsin(sec \ x))) = sin\left(-$\frac{\pi }{2}\right) {}$

and

$x = 0, \pi$ are the two solutions.

Calculate .

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The arcsecant function takes a trigonometric ratio on the unit circle as its input and results in an angle measure as its output. The given function can therefore be rewritten as

$Arcsec2=Arccos$\frac{1}{2}$$

and is the angle measure $\Theta$ which, when applied to the cosine function $cos\Theta$ , results in $\frac{1}{2}$ . Notice that the arcsecant function as expressed in the statement of the problem is capitalized; hence, we are looking for the "principal" angle measure, or the one which lies between and $\pi rad$ . Since $cos($\frac{\pi}$3)=\frac{1}{2}$$ , and since $\frac{\pi}{3}$ lies between and $\pi rad$ ,

$Arcsec2=\frac{\pi}{3}$$ .

Calculate .

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The domain on the argument for is

$-$\frac{\pi}{2}$<x<$\frac{\pi}{2}$$ .

The range of the function is not defined at $-$\frac{\pi}{2}$$ or $\frac{\pi}{2}$ , and so the domain of its inverse, , does not include those values. Hence, we must find the angle between $-$\frac{\pi}{2}$$ and $\frac{\pi}{2}$ for which .

Since $$\frac{sinx}{cosx}$=tanx$ , the equation can be rewritten as

$$\frac{sinx}{cosx}$=1$ ,

or

for some x between $-$\frac{\pi}{2}$$ and $\frac{\pi}{2}$ .

Now, when $x=\frac{\pi}{4}$$ , since $sin($\frac{\pi}{4}$)=cos($\frac{\pi}{4}$)=\frac{\sqrt{2}$}{2}$ .

Therefore,

$arctan1=\frac{\pi}{4}$$ .

What is $\theta$ if and ?

Tap to see back →

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$$

$\tan=\frac{10}{8}$$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow $51.34^{\circ}$$

If you calculated the angle's measure to be then your calculator was set to radians and needs to be set on degrees.

Which of the following is the degree equivalent of the inverse trigonometric function

$\arccos x=\frac{1}{2}$$ ?

Tap to see back →

The $\arccos$ is the reversal of the cosine function. That means that if $\cos \theta=y$ , then $\arccos y=\theta$ .

Therefore,

$\theta=\arccos\left($\frac{1}{2}\right)=60^\circ$

For the above triangle, what is $\theta$ if and ?

Tap to see back →

We need to use a trigonometric function to find $\theta$ . We are given the opposite and adjacent sides, so we can use the $\tan$ and $\arctan$ functions.

$\tan(\theta) = $\frac{\textup{opposite}$}{\textup{adjacent}}$

$\tan(\theta) = $\frac{20}{14}$$

$\theta = \arctan\left ( $\frac{20}{14}$ \right )$

$\theta = $55.0^{\circ}$$

For the above triangle, what is $\theta$ if and ?

Tap to see back →

We need to use a trigonometric function to find $\theta$ . We are given the opposite and hypotenuse sides, so we can use the $\sin$ and $\arcsin$ functions.

$\sin(\theta) = $\frac{\textup{opposite}$}{\textup{hypotenuse}}$

$\sin(\theta) = $\frac{17}{25}$$

$\theta = \arcsin\left ( $\frac{17}{25}$ \right )$

$\theta = $42.8^{\circ}$$

Assuming the angle in degrees, determine the value of .

Tap to see back →

To evaluate , it is necessary to know the existing domain and range for these inverse functions.

Inverse sine:

$R \textup{ (in radians)}: [-$\frac{\pi}{2}$,$\frac{\pi}{2}$]$

Inverse cosine:

$R\textup{ (in radians)}: [0,\pi]$

Inverse tangent:

$D: \textup{All real numbers.}$

$R\textup{ (in radians)}: (-$\frac{\pi}{2}$,$\frac{\pi}{2}$)$

Evaluate each term. The final answers must return an angle.

If

$cos(arcsin(sec \ x)) = 0{}$ ,

what value(s) does take?

Assume that $0\leq x< 2\pi$

Tap to see back →

If $cos(arcsin(sec \ x)) = 0{}$ , then we can apply the cosine inverse to both sides:

$arcsin(sec \ x)) = $\frac{\pi }{2}$ \ or -$\frac{\pi }{2}$$

Since cosine and cosine inverse undo each other; we can then apply sine and secant inverse functions to obtain the solution.

${}sin(arcsin(sec \ x))) = sin\left($\frac{\pi }{2}$ \right)$ and $sin(arcsin(sec \ x))) = sin\left(-$\frac{\pi }{2}\right) {}$

and

$x = 0, \pi$ are the two solutions.

Calculate .

Tap to see back →

The arcsecant function takes a trigonometric ratio on the unit circle as its input and results in an angle measure as its output. The given function can therefore be rewritten as

$Arcsec2=Arccos$\frac{1}{2}$$

and is the angle measure $\Theta$ which, when applied to the cosine function $cos\Theta$ , results in $\frac{1}{2}$ . Notice that the arcsecant function as expressed in the statement of the problem is capitalized; hence, we are looking for the "principal" angle measure, or the one which lies between and $\pi rad$ . Since $cos($\frac{\pi}$3)=\frac{1}{2}$$ , and since $\frac{\pi}{3}$ lies between and $\pi rad$ ,

$Arcsec2=\frac{\pi}{3}$$ .

Calculate .

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The domain on the argument for is

$-$\frac{\pi}{2}$<x<$\frac{\pi}{2}$$ .

The range of the function is not defined at $-$\frac{\pi}{2}$$ or $\frac{\pi}{2}$ , and so the domain of its inverse, , does not include those values. Hence, we must find the angle between $-$\frac{\pi}{2}$$ and $\frac{\pi}{2}$ for which .

Since $$\frac{sinx}{cosx}$=tanx$ , the equation can be rewritten as

$$\frac{sinx}{cosx}$=1$ ,

or

for some x between $-$\frac{\pi}{2}$$ and $\frac{\pi}{2}$ .

Now, when $x=\frac{\pi}{4}$$ , since $sin($\frac{\pi}{4}$)=cos($\frac{\pi}{4}$)=\frac{\sqrt{2}$}{2}$ .

Therefore,

$arctan1=\frac{\pi}{4}$$ .

What is $\theta$ if and ?

Tap to see back →

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$$

$\tan=\frac{10}{8}$$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow $51.34^{\circ}$$

If you calculated the angle's measure to be then your calculator was set to radians and needs to be set on degrees.

Which of the following is the degree equivalent of the inverse trigonometric function

$\arccos x=\frac{1}{2}$$ ?

Tap to see back →

The $\arccos$ is the reversal of the cosine function. That means that if $\cos \theta=y$ , then $\arccos y=\theta$ .

Therefore,

$\theta=\arccos\left($\frac{1}{2}\right)=60^\circ$

For the above triangle, what is $\theta$ if and ?

Tap to see back →

We need to use a trigonometric function to find $\theta$ . We are given the opposite and adjacent sides, so we can use the $\tan$ and $\arctan$ functions.

$\tan(\theta) = $\frac{\textup{opposite}$}{\textup{adjacent}}$

$\tan(\theta) = $\frac{20}{14}$$

$\theta = \arctan\left ( $\frac{20}{14}$ \right )$

$\theta = $55.0^{\circ}$$

For the above triangle, what is $\theta$ if and ?

Tap to see back →

We need to use a trigonometric function to find $\theta$ . We are given the opposite and hypotenuse sides, so we can use the $\sin$ and $\arcsin$ functions.

$\sin(\theta) = $\frac{\textup{opposite}$}{\textup{hypotenuse}}$

$\sin(\theta) = $\frac{17}{25}$$

$\theta = \arcsin\left ( $\frac{17}{25}$ \right )$

$\theta = $42.8^{\circ}$$