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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 #1 : Product Of Sines And Cosines

Use the product of cosines to evaluate $cos(\frac{\pi}{6})cos(\frac{5\pi}{3})$

Possible Answers:

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

$4\pi$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{3\pi}{4}$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\sqrt{3}$

Correct answer:

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

Explanation:

We are using the identity $cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$ . We will let $\alpha = \frac{\pi}{6}$ and $\beta = \frac{5\pi}{3}$ .

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) = \frac{1}{2}[cos(\frac{\pi}{6} + \frac{5\pi}{3}) + cos(\frac{\pi}{6} - \frac{5\pi}{3})]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[cos(\frac{11\pi}{6}) + cos(\frac{-9\pi}{6})]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[\frac{\sqrt{3}}{2} +0]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

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Example Question #1 : Product Of Sines And Cosines

Use the product of sines to evaluate $sin(\frac{3x}{4})sin(\frac{6x}{2})$ where $x = \frac{pi}{3}$

Possible Answers:

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{\sqrt{2}}{2}$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =\frac{\sqrt{3}}{2}$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =0$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =\sqrt{2}$

Correct answer:

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =0$

Explanation:

The formula for the product of sines is $sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha-\beta) - cos(\alpha + \beta)$ . We will let $\alpha = \frac{3x}{4}$ and $\beta = \frac{6x}{2}$ .

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{3x}{4} - \frac{6x}{2}) - cos(\frac{3x}{4} + \frac{6x}{2})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9x}{4}) - cos(\frac{15x}{4})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9}{4}\frac{\pi}{3}) - cos(\frac{15}{4}\frac{\pi}{3})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9\pi}{12}) - cos(\frac{15\pi}{12})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-3\pi}{4}) - cos(\frac{5\pi}{4})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{-\sqrt{2}}{2}]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[0]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = 0$

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Example Question #1 : Product Of Sines And Cosines

True or False: All of the product-to-sum identities can be obtained from the sum-to-product identities

Possible Answers:

False

True

Correct answer:

True

Explanation:

All of these identities are able to be obtained by the sum-to-product identities by either adding or subtracting two of the sum identities and canceling terms. Through some algebra and manipulation, you are able to derive each product identity.

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Example Question #2 : Product Of Sines And Cosines

Use the product of sine and cosine to evaluate $sin(\frac{\pi}{4})cos(\pi)$ .

Possible Answers:

$sin(\frac{\pi}{4]cos(\pi) = -\sqrt{2}$

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

$sin(\frac{\pi}{4]cos(\pi) = \frac{\sqrt{2}}{2}$

$sin(\frac{\pi}{4]cos(\pi) = \sqrt{2}$

Correct answer:

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

Explanation:

The identity that we will need to utilize to solve this problem is $sin(\alpha)cos(\beta) = \frac{1}{2}[sin(\alpha + \beta) + sin(\alpha - \beta)]$ . We will let $\alpha = \frac{\pi}{4}$ and $\beta = \pi$ .

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{\pi}{4} + \pi) +sin(\frac{\pi}{4} - \pi)]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{5\pi}{4} + sin(\frac{-3\pi}{4}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{\sqrt{2}}{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-2\sqrt{2}}{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[-\sqrt{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

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Example Question #2 : Product Of Sines And Cosines

Use the product of cosines to evaluate cos(4x)cos(2x) . Keep your answer in terms of .

Possible Answers:

$cos(4x)cos(2x) =\frac{1}{2}[cos(8x)]$

cos(4x)cos(2x) =cos(6x) + cos(2x)

cos(4x)cos(2x) =cos(3x) + cos(x)

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

Correct answer:

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

Explanation:

The identity we will be using is $cos(\alpha)cos(\beta) = \frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$ . We will let $\alpha = 4x$ and $\beta = 2x$ . $cos(4x)cos(2x) = \frac{1}{2}[cos(4x + 2x) + cos(4x - 2x)]$

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

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Example Question #2 : Product Of Sines And Cosines

Use the product of sines to evaluate sin(45)sin(30) .

Possible Answers:

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

sin(45)sin(30) = 2

sin(45)sin(30) = 4

$sin(45)sin(30) = \sqrt{2}$

Correct answer:

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

Explanation:

The identity that we will need to use is $sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

$sin(45)sin(30) = \frac{1}{2}[cos(45 - 30) - cos(45 + 30)]$

$sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{4} - \frac{\pi}{6}) - cos(\frac{\pi}{4} + \frac{\pi}{6})]$

$sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{12}) - cos(\frac{5\pi}{12})]$

$sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{2}}{4}(\sqrt{3} +1) - \frac{\sqrt{2}}{4}(\sqrt{3} -1)]$

$sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{6}+\sqrt{2}}{4} - \frac{\sqrt{6}-\sqrt{2}}{4}]$

$sin(45)sin(30) = \frac{1}{2}[\frac{2\sqrt{2}}{4}]$

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

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Example Question #1 : Graphs Of Inverse Trigonometric Functions

True or False: The inverse of the function $\sin (x)$ is also a function.

Possible Answers:

True

False

Correct answer:

False

Explanation:

Consider the graph of the function sin(x) . It passes the vertical line test, that is if a vertical line is drawn anywhere on the graph it only passes through a single point of the function. This means that sin(x) is a function.

Screen shot 2020 08 27 at 11.59.46 am

Now, for its inverse to also be a function it must pass the horizontal line test. This means that if a horizontal line is drawn anywhere on the graph it will only pass through one point.

Screen shot 2020 08 27 at 12.00.18 pm

This is not true, and we can also see that if we graph the inverse of sin(x) ( $sin^{-1}(x)$ ) that this does not pass the vertical line test and therefore is not a function. If you wish to graph the inverse of sin(x) , then you must restrict the domain so that your graph will pass the vertical line test.

Screen shot 2020 08 27 at 12.00.40 pm

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Example Question #1 : Graphs Of Inverse Trigonometric Functions

Which of the following is the graph of the inverse of sin(x) with $D: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ ?

Possible Answers:

Screen shot 2020 08 27 at 10.47.13 am

Screen shot 2020 08 27 at 10.47.06 am

Screen shot 2020 08 27 at 10.46.15 am

Screen shot 2020 08 27 at 10.45.10 am

Correct answer:

Screen shot 2020 08 27 at 10.46.15 am

Explanation:

Note that the inverse of sin(x) is not $\frac{1}{sin(x)}$ , that is the reciprocal. The inverse of sin(x) is $sin^{-1}(x)$ also written as arcsin(x) . The graph of sin(x) with $D: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ is as follows.

Screen shot 2020 08 27 at 10.45.10 am

And so the inverse of this graph must be the following with $D: -1 \leq x \leq 1$ and $R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

Screen shot 2020 08 27 at 10.46.15 am

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Example Question #3 : Graphs Of Inverse Trigonometric Functions

Which best describes the easiest method to graph an inverse trigonometric function (or any function) based on the parent function?

Possible Answers:

The values are swapped with $\frac{1}{y}$ so where y = cos(x) for the parent function, $x = cos(\frac{1}{y})$ for the inverse function.

Let $y = \frac{1}{y}$ so where y = cos(x) for the parent function $y = \frac{1}{cos(x)}$ for the inverse function.

The and values are switched so where y = cos(x) for the parent function, x = cos(y) for the inverse function.

Let $x = \frac{1}{x}$ so where y = cos(x) for the parent function, $y = cos(\frac{1}{x})$ for the inverse function.

Correct answer:

The and values are switched so where y = cos(x) for the parent function, x = cos(y) for the inverse function.

Explanation:

To find an inverse function you swap the and values. Take cos(x) for example, to find the inverse we use the following method.

y = cos(x)

x = cos(y) (swap the and values)

$cos^{-1}(x) = y$ (solving for )

$y = cos^{-1}(x)$

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Example Question #4 : Graphs Of Inverse Trigonometric Functions

Which of the following represents the graph of $tan^{-1}(x)$ with $R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$ ?

Possible Answers:

Screen shot 2020 08 28 at 9.02.30 am

Screen shot 2020 08 28 at 9.00.48 am

Screen shot 2020 08 28 at 9.02.37 am

Screen shot 2020 08 28 at 9.01.00 am

Correct answer:

Screen shot 2020 08 28 at 9.01.00 am

Explanation:

If we are looking for the graph of $tan^{-1}(x)$ with $R: \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ , that means this is the inverse of tan(x) with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ . The graph of tan(x) with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ is

Screen shot 2020 08 28 at 9.00.48 am

Switching the and values to graph the inverse we get the graph

Screen shot 2020 08 28 at 9.01.00 am

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #2 : Product Of Sines And Cosines

Example Question #2 : Product Of Sines And Cosines

Example Question #2 : Product Of Sines And Cosines

Example Question #1 : Graphs Of Inverse Trigonometric Functions

Example Question #1 : Graphs Of Inverse Trigonometric Functions

Example Question #3 : Graphs Of Inverse Trigonometric Functions

Example Question #4 : Graphs Of Inverse Trigonometric Functions

All Trigonometry Resources

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