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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Arcsin, Arccos, Arctan

Soh_cah_toa

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

Possible Answers:

$45.6^{\circ}$

$60.0^{\circ}$

$44.4^{\circ}$

$55.0^{\circ}$

$35.0^{\circ}$

Correct answer:

$55.0^{\circ}$

Explanation:

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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Example Question #2 : Arcsin, Arccos, Arctan

Soh_cah_toa

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

Possible Answers:

$47.2^{\circ}$

$11.8^{\circ}$

$34.2^{\circ}$

$55.8^{\circ}$

$42.8^{\circ}$

Correct answer:

$42.8^{\circ}$

Explanation:

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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Example Question #2 : Arcsin, Arccos, Arctan

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

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

Possible Answers:

$360^\circ$

$60^\circ$

$1^\circ$

$2^\circ$

$30^\circ$

Correct answer:

$60^\circ$

Explanation:

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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Example Question #3 : Arcsin, Arccos, Arctan

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

Possible Answers:

$\textup{The answer cannot be determined.}$

Correct answer:

Explanation:

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:

D: [-1,1]

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

Inverse cosine:

D: [-1,1]

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

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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Example Question #3 : Arcsin, Arccos, Arctan

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

what value(s) does take?

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

Possible Answers:

$\frac{\pi}{2}, -\frac{\pi}{2}$

$0, \pi$

$\frac{\pi}{2}$

No real solution.

Correct answer:

$0, \pi$

Explanation:

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

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

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

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Example Question #1 : Arcsin, Arccos, Arctan

Calculate Arcsec(2) .

Possible Answers:

$\frac{\pi}{3} and \frac{3\pi}{2}$

$\frac{\pi}{3}$ and $\frac{5\pi}{3}$

$\frac{3\pi}{2}$

$\frac{5\pi}{3}$

$\frac{\pi}{3}$

Correct answer:

$\frac{\pi}{3}$

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

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

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 0rad and $\pi rad$ . Since $cos(\frac{\pi}3)=\frac{1}{2}$ , and since $\frac{\pi}{3}$ lies between 0rad and $\pi rad$ ,

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

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Example Question #1 : Arcsin, Arccos, Arctan

Calculate arctan1 .

Possible Answers:

$\frac{\pi}{2}$

$-\frac{\pi}{2}$

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{4}$

Explanation:

The domain on the argument for is

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

The range of the function tanx 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 tanx=1 .

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

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

sinx=cosx

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

Now, sinx=cosx 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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Example Question #1 : Angular Velocity

What is the formula for angular velocity?

Possible Answers:

$\omega = \frac{r}{l_A}$

$\omega = \frac{\theta}{r}$

$\omega = \frac{\theta}{t}$

$\omega = \frac{\theta}{l_A}$

Correct answer:

$\omega = \frac{\theta}{t}$

Explanation:

Angular velocity is a measure of the time it takes to travel over a certain arc that is formed by a central angle. $\omega$ is usually used to represent the angular velocity, $\theta$ is the measure of the central angle, and is the time it takes to travel from point A to B. This formula looks similar to the measure of an arc, $\theta = \frac{r}{l_A}$ , but we are considering the velocity in terms of the measure of the angle over the amount of time it took to cover that distance.

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Example Question #2 : Angular Velocity

What is the angular velocity of an angle whose measure is $\frac{5\pi}{4}$ when it took a point 6 seconds to rotate through the angle.

Possible Answers:

$\omega = \frac{5\pi}{24}$ radians/seconds

$\omega = \frac{5\pi}{20}$ radians/seconds

$\omega = \frac{10\pi}{24}$ radians/seconds

$\omega = 5\pi$ radians/seconds

Correct answer:

$\omega = \frac{5\pi}{24}$ radians/seconds

Explanation:

We have enough information to plug into the angular velocity formula to solve this problem. $\theta = \frac{5\pi}{4}$ and t = 6 seconds.

$\omega = \frac{\theta}{t}$

$\omega = (\frac{5\pi}{4})(\frac{1}{6})$

$\omega = \frac{5\pi}{24}$

So our answer is $\omega = \frac{5\pi}{24}$ radians/seconds

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Example Question #3 : Angular Velocity

True or False: The angular velocity is a measure of an angle of rotation over time.

Possible Answers:

False

True

Correct answer:

True

Explanation:

consider the figure below. Think of line A being connected to a hinge where $\theta$ is. We are measuring the amount of time it takes for line A to rotate and align with line B. This is a measure of angle rotation over time.

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #2 : Arcsin, Arccos, Arctan

Example Question #2 : Arcsin, Arccos, Arctan

Example Question #3 : Arcsin, Arccos, Arctan

Example Question #3 : Arcsin, Arccos, Arctan

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #1 : Angular Velocity

Example Question #2 : Angular Velocity

Example Question #3 : Angular Velocity

All Trigonometry Resources

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