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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Using The Unit Circle

What is $\cos(\pi)$ ?

Possible Answers:

$\frac{\sqrt3}{2}$

$\frac{\sqrt2}{2}$

Correct answer:

Explanation:

If you examine the unit circle, you'll see that the value of $\cos(\pi)=-1$ . You can also get this by examining a cosine graph and you'll see it crosses the point $(\pi,-1)$ .

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Example Question #81 : Pre Calculus

Which one of these is positive in quadrant III?

Possible Answers:

Cosine

All trig functions

Tangent

No trig functions

Sine

Correct answer:

Tangent

Explanation:

The pattern for positive functions is All Student Take Calculus. In quandrant I, all trigonometric functions are positive. In quadrant II, sine is positive. In qudrant III, tangent is positive. In quadrant IV, cosine is positive.

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Example Question #1 : Understanding Coterminal Angles

Find a coterminal angle for $-55^{\circ}$ .

Possible Answers:

$-15^{\circ}$

$305^{\circ}$

$360^{\circ}$

$300^{\circ}$

$85^{\circ}$

Correct answer:

$305^{\circ}$

Explanation:

Coterminal angles are angles that, when drawn in the standard position, share a terminal side. You can find these angles by adding or subtracting 360 to the given angle. Thus, the only angle measurement that works from the answers given is $305^{\circ}$ .

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Example Question #2 : Understanding Coterminal Angles

Which of the following angles is coterminal with $\frac{5\pi}{9}$ ?

Possible Answers:

$\frac{41\pi}{9}$

$\frac{23\pi}{9}$

$-\frac{13\pi}{9}$

Each angle given in the other choices is coterminal with $\frac{5\pi}{9}$ .

$-\frac{31\pi}{9}$

Correct answer:

Each angle given in the other choices is coterminal with $\frac{5\pi}{9}$ .

Explanation:

For an angle to be coterminal with $\theta$ , that angle must be of the form $\theta + 2\pi N$ for some integer - or, equivalently, the difference of the angle measures multiplied by $\frac{1}{2\pi}$ must be an integer. We apply this test to all four choices.

$\frac{41\pi}{9}$ :

$\left ( \frac{1}{2\pi } \right ) \left (\frac{41\pi}{9}- \frac{5\pi}{9}\right )$

$=\left ( \frac{1}{2\pi } \right )\left ( \frac{36\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )4 \pi =2$

$\frac{23\pi}{9}$ :

$\left ( \frac{1}{2\pi } \right ) \left (\frac{23\pi}{9}- \frac{5\pi}{9}\right )$

$=\left ( \frac{1}{2\pi } \right )\left ( \frac{18\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )2 \pi =1$

$-\frac{13\pi}{9}$ :

$\left ( \frac{1}{2\pi } \right ) \left (-\frac{13\pi}{9}- \frac{5\pi}{9}\right )$

$=\left ( \frac{1}{2\pi } \right )\left ( \frac{-18\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )\left (- 2 \pi \right )=-1$

$-\frac{31\pi}{9}$ :

$\left ( \frac{1}{2\pi } \right ) \left (-\frac{31\pi}{9}- \frac{5\pi}{9}\right )$

$=\left ( \frac{1}{2\pi } \right )\left ( \frac{-36\pi}{9} \right ) =\left ( \frac{1}{2\pi } \right )\left (- 4 \pi \right )=-2$

All four choices pass the test, so all four angles are coterminal with $\frac{5\pi}{9}$ .

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Example Question #1 : Trigonometry

$\sin(x^\circ)=\frac{1}{2}$

What is ?

Possible Answers:

$60^\circ$

$0.009^\circ$

$26.57^\circ$

$1^\circ$

$30^\circ$

Correct answer:

$30^\circ$

Explanation:

To get rid of $\sin(x^\circ)$ , we take the $\arcsin$ or $\sin^{-1}$ of both sides.

$\sin(x^\circ)=\frac{1}{2}$

$\sin^{-1}(\sin(x^\circ))=\sin^{-1}(\frac{1}{2})$

$x^\circ=\sin^{-1}(\frac{1}{2})$

$x^\circ=30^\circ$

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Example Question #2 : Trigonometric Operations

Trig_id

$o=7.6, h=15.7, \Theta = ?$

Possible Answers:

$59.28^{\circ}$

$50.53^{\circ}$

$0.50^{\circ}$

$28.95^{\circ}$

$85.92^{\circ}$

Correct answer:

$28.95^{\circ}$

Explanation:

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

$\sin=\frac{opposite}{hypotenuse}$

$\sin=\frac{7.6}{15.7}$

$\sin=0.4841$

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

$\theta=\arcsin0.4841$

$\rightarrow 28.95^{\circ}$

If you calculated the angle's measure to be $\dpi{100} 0.50^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Example Question #82 : Pre Calculus

What is the amplitude of the following equation?

$f(x)=4\sin{3x}+2$

Possible Answers:

Correct answer:

Explanation:

Based on the generic form $a\sin{bx}+c$ , a is the amplitude. Thus, the amplitude is 4.

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Example Question #24 : Graphing The Sine And Cosine Functions

What is one possible length of side if right triangle ABC has $\angle A = 30^{\circ}$ and side AC = 6 ?

(Hint: There are two possible answers, but only one of them is listed.)

Possible Answers:

$3\sqrt{2}$

$3\sqrt{3}$

Correct answer:

$3\sqrt{3}$

Explanation:

First we must set up our equation given the information.

$\cos 30 = \frac{AB}{6}$

$AB = 6\cos30$

$AB=6*\frac{\sqrt3}{2}$

$AB = 3\sqrt{3}$

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Example Question #31 : Graphing The Sine And Cosine Functions

How many -intercepts does the function $y=sin(2 \pi x)$ have in the domain [0,5] ?

Possible Answers:

Correct answer:

Explanation:

The period of the sine function is $2 \pi$ . The period of our new function is 1. Each period will have two zeroes and there will be one tacked on at the end when the domain is closed. There will be -intercepts at $x=0,\frac{1}{2},1,\frac{3}{2},2\frac{5}{2},3,\frac{7}{2},4,\frac{9}{2},\textup{and}\ 5$ .

The graph is below.

Wolframalpha--graph_of_y__sin2pix_from_x0_to_x5--2014-12-18_2321

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Example Question #85 : Pre Calculus

If $\sec x < 0$ and $\cot x > 0$ , then which of the following must be true about ?

Possible Answers:

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

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

$-\frac{\pi}{2} < x < 0$

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

$0 < x < \frac{\pi}{2}$

Correct answer:

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

Explanation:

It is a question of what quadrant is in.

A negative value for secant indicates quadrant II or III. Since secant is the reciprocal of cosine, the measurement includes the x value and the r value with regards to position.To get a negative value for secant or cosine we will need a negative x value and either a positive or negative y value to get the correct r value.

A positive value for cotangent indicates quadrant I or III. Since cotangent is the reciprocal of tangent, the measurement includes the x and y values with regards to the position. To get a cotangent that is positive we will need a positive x value and either a positive or negative y value.

The overlap between these two statements is quadrant III. Therefore, must be in quadrant III.

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #6 : Using The Unit Circle

Example Question #81 : Pre Calculus

Example Question #1 : Understanding Coterminal Angles

Example Question #2 : Understanding Coterminal Angles

Example Question #1 : Trigonometry

Example Question #2 : Trigonometric Operations

Example Question #82 : Pre Calculus

Example Question #24 : Graphing The Sine And Cosine Functions

Example Question #31 : Graphing The Sine And Cosine Functions

Example Question #85 : Pre Calculus

All Precalculus Resources

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