Arcsin, Arccos, Arctan - Trigonometry

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Question

What is $\theta$ if and ?

Answer

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 $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Question

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

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

Answer

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

Therefore,

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

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Question

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

Answer

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.

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

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

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

$\theta = 55.0^{\circ}$

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Question

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

Answer

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.

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

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

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

$\theta = 42.8^{\circ}$

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Question

Assuming the angle in degrees, determine the value of $sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)$ .

Answer

To evaluate $sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)$ , 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.

$sin^{-1}(1)=\frac{\pi}{2} \textup{ radians}=90 \textup{ degrees}$

$cos^{-1}(1)=0 \textup{ degrees}$

$tan^{-1}(1)=\frac{\pi}{4} \textup{ radians}= 45 \textup{ degrees}$

$sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)=90-0-45=45 \textup{ degrees}$

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Question

If

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

what value(s) does take?

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

Answer

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

$cos^{-1}(cos(arcsin(sec \ x))) = cos^{-1}(0)$

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

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.

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Question

Calculate .

Answer

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 $\pi rad$ ,

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

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Question

Calculate .

Answer

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,

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

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