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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

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

Example Questions

Example Question #6 : Graphing Tangent And Cotangent

Which of the following is the graph of cot(x) ?

Possible Answers:

Screen shot 2020 08 27 at 2.16.51 pm

Screen shot 2020 08 27 at 2.18.25 pm

Screen shot 2020 08 27 at 2.16.42 pm

Screen shot 2020 08 27 at 2.18.30 pm

Correct answer:

Screen shot 2020 08 27 at 2.16.51 pm

Explanation:

To derive the graph of cot(x) recall that $cot(x) = \frac{1}{tan(x)}$ . So the tangent and cotangent graphs are reciprocals of one another. We will consider the tangent graph since it is one we are more familiar with:

Screen shot 2020 08 27 at 2.16.42 pm

Now we will simply invert the tangent graph to get the cotangent graph

Screen shot 2020 08 27 at 2.16.46 pm

And we are left with our cotangent graph

Screen shot 2020 08 27 at 2.16.51 pm

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Example Question #7 : Graphing Tangent And Cotangent

Which of the following is the graph of $4cot(x + \frac{2\pi}{5})$ .

Possible Answers:

Screen shot 2020 08 27 at 11.12.03 am

Screen shot 2020 08 27 at 11.12.10 am

Screen shot 2020 08 27 at 11.11.11 am

Screen shot 2020 08 27 at 11.11.56 am

Correct answer:

Screen shot 2020 08 27 at 11.11.11 am

Explanation:

First, we will consider the graph of cot(x) and apply transformations step-by-step. The graph of cot(x) is

Screen shot 2020 08 27 at 11.08.57 am

The general form of a cotangent transformation function is D + Acot(B(x - C)) . For our function $4cot(x + \frac{2\pi}{5})$ , A = 4 and so we need to increase the amplitude 4 units.

Screen shot 2020 08 27 at 11.09.39 am

B = 1 here so we do not need to make any changes to the period of this graph. $C = -\frac{2\pi}{5}$ , giving us a negative phase shift of $\frac{2\pi}{5}$ units.

Screen shot 2020 08 27 at 11.10.16 am

This leaves us with our graph of $4cot(x + \frac{2\pi}{5})$ .

Screen shot 2020 08 27 at 11.11.11 am

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Example Question #1 : Graphing Tangent And Cotangent

True or False: The period of tangent and cotangent function is $\pi$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is because $tan(x) = \frac{sin(x)}{cos(x)}$ and $cos(n\pi) = 0$ causing the tangent function to be undefined at these points and forming a vertical asymptote. This is also true for the cotangent function because $cot(x) = \frac{1}{tan(x)}$ so wherever tan(x) is zero or undefined, cotangent will be as well.

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Example Question #1 : Understanding Trigonometric Equations

A triangle has sides , , of lengths , , respectively. The angle opposite each side is called , , , respectively. The sine of which angle and the cosine and which different angle will both yield $\frac{8}{17}$

In the answer, list the sine first and the cosine second.

Possible Answers:

a, b

a, c

c, a

b, c

b, a

Correct answer:

a, b

Explanation:

Screen_shot_2015-03-07_at_2.36.12_pm

This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of and the cosine of will yield the correct answer.

$\sin(a)=\frac{opp}{hyp}=\frac{8}{17}$

$\cos(b)=\frac{adj}{hyp}=\frac{8}{17}$

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Example Question #2 : Understanding Trigonometric Equations

Solve the equation over the domain 0 < x < 360 (answer in degrees).

3sin(x)-2 = 1

Possible Answers:

0, 360

180

270

Correct answer:

Explanation:

Rearrange algebraically so that,

3sin(x)=3

sinx = 1 .

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

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Example Question #2 : Trigonometric Equations

Solve each equation over the domain 0 < x < 360 (answer in degrees).

tan(x-15) = 1

Possible Answers:

315, 135

45, 225

60, 240

60, 150

Correct answer:

60, 240

Explanation:

First, think of the angle values for which tanx = 1 . (This is the equivalent of taking the arctan.)

$x-15=tan^{-1}(1)$

$x=tan^{-1}+15$ .

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240.

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Example Question #161 : Trigonometry

Solve each equation over the domain 0 < x < 360 (answer in degrees).

sec(x+7)-2 = 0

Possible Answers:

60, 300

30, 150

23, 143

53, 293

Correct answer:

53, 293

Explanation:

Rearrange the equation so that,

sec(x+7) = 2 .

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293.

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Example Question #1 : Setting Up Trigonometric Equations

Solve each equation over the interval 0 < x < $2\pi$

4cos^2(x) - 2 = 0

Possible Answers:

$\frac{\pi}{6}, \frac{7\pi}{6}$

$\frac{\pi}{4}, \frac{7\pi}{4}$

$\frac{3\pi}{4}, \frac{5\pi}{4}$

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

Correct answer:

$\frac{\pi}{4}, \frac{7\pi}{4}$

Explanation:

Rearrange the equation so that,

$cos^2x = \frac{1}{2}$ .

Take the square of both sides and then recall the angle measures for which,

$cos(x)={\frac{1}{\sqrt2}}$ .

These measures over the interval $0<x< 2 \pi$

$x=\frac{\pi}{4}; \frac{7\pi}{4}$ .

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Example Question #2 : Setting Up Trigonometric Equations

Solve each quation over the interval $0 < x < 2\pi$

sin^2(x) - 1 = 0

Possible Answers:

$0, \: 2\pi$

$0, \frac{\pi}{2}$

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

$\frac{\pi}{4}, \frac{7\pi}{4}$

Correct answer:

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

Explanation:

Rearrange the equation so that,

sin^2x =1 .

Take the square of both sides and find the angles for which

$sin(x)=\pm1$ .

These two angles are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

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Example Question #2 : Trigonometric Equations

Solve for using trigonometric ratios.

Set up trig 1

Possible Answers:

$6\sqrt2$

$\sqrt{12}$

$\frac{1}{\sqrt2}$

Correct answer:

$6\sqrt2$

Explanation:

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

$sin\left(\frac{\pi}{4}\right)= \frac{6}{x}$

The sine of $\frac{\pi}{4}$ is $\frac{1}{\sqrt2}$ , so we can substitue that in:

$\frac{1}{\sqrt2}=\frac{6}{x}$

cross multiplying gives us $x=6\sqrt2$ .

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #6 : Graphing Tangent And Cotangent

Example Question #7 : Graphing Tangent And Cotangent

Example Question #1 : Graphing Tangent And Cotangent

Example Question #1 : Understanding Trigonometric Equations

Example Question #2 : Understanding Trigonometric Equations

Example Question #2 : Trigonometric Equations

Example Question #161 : Trigonometry

Example Question #1 : Setting Up Trigonometric Equations

Example Question #2 : Setting Up Trigonometric Equations

Example Question #2 : Trigonometric Equations

All Trigonometry Resources

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