Trigonometry : Arcsin, Arccos, Arctan

Study concepts, example questions & explanations for Trigonometry

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Example Questions

Example Question #1 : Arcsin, Arccos, Arctan

Trig_id

What is \(\displaystyle \theta\) if \(\displaystyle o=10\) and \(\displaystyle a=8\)?

Possible Answers:

\(\displaystyle 53.14^{\circ}\)

\(\displaystyle 51.34^{\circ}\)

\(\displaystyle 55.14^{\circ}\)

\(\displaystyle 0.90^{\circ}\)

\(\displaystyle 89.60^{\circ}\)

Correct answer:

\(\displaystyle 51.34^{\circ}\)

Explanation:

In order to find \(\displaystyle \theta\) 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 \(\displaystyle \tan\) of \(\displaystyle \theta\) and its measure in degrees by utilizing the \(\displaystyle \arctan\) function.

\(\displaystyle \tan=\frac{opposite}{adjacent}\)

\(\displaystyle \tan=\frac{10}{8}\)

\(\displaystyle \tan=1.25\)

Now to find the measure of the angle using the \(\displaystyle \arctan\) function.

\(\displaystyle \theta=\arctan1.25\)

\(\displaystyle \rightarrow 51.34^{\circ}\)

If you calculated the angle's measure to be \(\displaystyle 0.90^{\circ}}\) then your calculator was set to radians and needs to be set on degrees.

Example Question #2 : Arcsin, Arccos, Arctan

Soh_cah_toa

For the above triangle, what is \(\displaystyle \theta\) if \(\displaystyle o = 20\) and \(\displaystyle a = 14\)?

Possible Answers:

\(\displaystyle 35.0^{\circ}\)

\(\displaystyle 60.0^{\circ}\)

\(\displaystyle 44.4^{\circ}\)

\(\displaystyle 45.6^{\circ}\)

\(\displaystyle 55.0^{\circ}\)

Correct answer:

\(\displaystyle 55.0^{\circ}\)

Explanation:

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

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

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

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

\(\displaystyle \theta = 55.0^{\circ}\)

Example Question #3 : Arcsin, Arccos, Arctan

Soh_cah_toa

For the above triangle, what is \(\displaystyle \theta\) if \(\displaystyle o = 17\) and \(\displaystyle h = 25\)?

Possible Answers:

\(\displaystyle 11.8^{\circ}\)

\(\displaystyle 47.2^{\circ}\)

\(\displaystyle 42.8^{\circ}\)

\(\displaystyle 34.2^{\circ}\)

\(\displaystyle 55.8^{\circ}\)

Correct answer:

\(\displaystyle 42.8^{\circ}\)

Explanation:

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

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

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

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

\(\displaystyle \theta = 42.8^{\circ}\)

Example Question #1 : Arcsin, Arccos, Arctan

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

\(\displaystyle \arccos x=\frac{1}{2}\)?

Possible Answers:

\(\displaystyle 30^\circ\)

\(\displaystyle 60^\circ\)

\(\displaystyle 2^\circ\)

\(\displaystyle 1^\circ\)

\(\displaystyle 360^\circ\)

Correct answer:

\(\displaystyle 60^\circ\)

Explanation:

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

Therefore,

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

Example Question #2 : Arcsin, Arccos, Arctan

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

Possible Answers:

\(\displaystyle \textup{The answer cannot be determined.}\)

\(\displaystyle 33\)

\(\displaystyle 57\)

\(\displaystyle 45\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 45\)

Explanation:

To evaluate \(\displaystyle sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)\), it is necessary to know the existing domain and range for these inverse functions.

Inverse sine:

\(\displaystyle D: [-1,1]\)

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

Inverse cosine:

\(\displaystyle D: [-1,1]\)

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

Inverse tangent:

\(\displaystyle D: \textup{All real numbers.}\)

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

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

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

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

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

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

Example Question #3 : Arcsin, Arccos, Arctan

If 

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

what value(s) does \(\displaystyle x\) take?

Assume that \(\displaystyle 0\leq x< 2\pi\)  

Possible Answers:

No real solution.

\(\displaystyle 0\)

\(\displaystyle \frac{\pi}{2}\)

\(\displaystyle \frac{\pi}{2}, -\frac{\pi}{2}\)

\(\displaystyle 0, \pi\)

Correct answer:

\(\displaystyle 0, \pi\)

Explanation:

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

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

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

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

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

\(\displaystyle sec(x) = 1\)                                          and            \(\displaystyle sec(x) = -1\)

\(\displaystyle x = 0, \pi\) are the two solutions. 

 

 

Example Question #2 : Arcsin, Arccos, Arctan

Calculate \(\displaystyle Arcsec(2)\).

Possible Answers:

\(\displaystyle \frac{\pi}{3}\) and \(\displaystyle \frac{5\pi}{3}\)

\(\displaystyle \frac{\pi}{3}\)

\(\displaystyle \frac{\pi}{3} and \frac{3\pi}{2}\)

\(\displaystyle \frac{5\pi}{3}\)

\(\displaystyle \frac{3\pi}{2}\)

Correct answer:

\(\displaystyle \frac{\pi}{3}\)

Explanation:

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 

\(\displaystyle Arcsec2=Arccos\frac{1}{2}\) 

and is the angle measure \(\displaystyle \Theta\) which, when applied to the cosine function \(\displaystyle cos\Theta\), results in \(\displaystyle \frac{1}{2}\). 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 \(\displaystyle 0rad\) and \(\displaystyle \pi rad\). Since \(\displaystyle cos(\frac{\pi}3)=\frac{1}{2}\), and since \(\displaystyle \frac{\pi}{3}\) lies between \(\displaystyle 0rad\) and \(\displaystyle \pi rad\),

\(\displaystyle Arcsec2=\frac{\pi}{3}\).

 

Example Question #3 : Arcsin, Arccos, Arctan

Calculate \(\displaystyle arctan1\).

Possible Answers:

\(\displaystyle -\frac{\pi}{4}\)

\(\displaystyle \frac{\pi}{2}\)

\(\displaystyle \frac{\pi}{4}\)

\(\displaystyle -\frac{\pi}{2}\)

Correct answer:

\(\displaystyle \frac{\pi}{4}\)

Explanation:

The domain on the argument \(\displaystyle x\) for \(\displaystyle arctanx\) is 

\(\displaystyle -\frac{\pi}{2}< x< \frac{\pi}{2}\) .

The range of the function \(\displaystyle tanx\) is not defined at \(\displaystyle -\frac{\pi}{2}\) or \(\displaystyle \frac{\pi}{2}\), and so the domain of its inverse, \(\displaystyle arctanx\), does not include those values. Hence, we must find the angle \(\displaystyle x\) between \(\displaystyle -\frac{\pi}{2}\) and \(\displaystyle \frac{\pi}{2}\) for which \(\displaystyle tanx=1\).

Since \(\displaystyle \frac{sinx}{cosx}=tanx\), the equation \(\displaystyle tanx=1\) can be rewritten as 

\(\displaystyle \frac{sinx}{cosx}=1\),

or

\(\displaystyle sinx=cosx\)

for some x between \(\displaystyle -\frac{\pi}{2}\) and \(\displaystyle \frac{\pi}{2}\).

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

Therefore, 

\(\displaystyle arctan1=\frac{\pi}{4}\).

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