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Common Core: High School - Functions : Trigonometric Functions

Study concepts, example questions & explanations for Common Core: High School - Functions

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All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\cos\left(\frac{7\pi}{6} \right )$

Possible Answers:

$\cos\left(\frac{7\pi}{6} \right )=-\frac{1}{2}$

$\cos\left(\frac{7\pi}{6} \right )=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{7\pi}{6} \right )=\frac{1}{2}$

$\cos\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{2}}{2}$

$\cos\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

Correct answer:

$\cos\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{7\pi}{6}$ .

$\theta = \frac{7\pi}{6}\Rightarrow \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{7\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)=\left(\cos\left(\frac{7\pi}{6}\right),\sin\left(\frac{7\pi}{6}\right)\right)$

$\cos\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}$ .

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Example Question #1 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\sin\left(\frac{7\pi}{6} \right )$

Possible Answers:

$\sin\left(\frac{7\pi}{6} \right )=-\frac{1}{\sqrt{2}}$

$\sin\left(\frac{7\pi}{6} \right )=\frac{1}{2}$

$\sin\left(\frac{7\pi}{6} \right )=-\frac{1}{2}$

$\sin\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

$\sin\left(\frac{7\pi}{6} \right )=\frac{1}{\sqrt{2}}$

Correct answer:

$\sin\left(\frac{7\pi}{6} \right )=-\frac{1}{2}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{7\pi}{6}$ .

$\theta = \frac{7\pi}{6}\Rightarrow \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2} \right )$

Step 3: Calculate the exact value of $\sin\left(\frac{7\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)=\left(\cos\left(\frac{7\pi}{6}\right),\sin\left(\frac{7\pi}{6}\right)\right)$

$\sin\left(\frac{7\pi}{6} \right )=-\frac{1}{2}$ .

Report an Error

Example Question #1 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\cos\left(\frac{11\pi}{6} \right )$

Possible Answers:

$\cos\left(\frac{11\pi}{6} \right )=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{11\pi}{6} \right )={\sqrt{3}}$

$\cos\left(\frac{11\pi}{6} \right )=-\frac{1}{2}$

$\cos\left(\frac{11\pi}{6} \right )=\frac{1}{2}$

$\cos\left(\frac{11\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

Correct answer:

$\cos\left(\frac{11\pi}{6} \right )=\frac{\sqrt{3}}{2}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{11\pi}{6}$ .

$\theta = \frac{11\pi}{6}\Rightarrow \left(\frac{\sqrt{3}}{2}, -\frac{1}{2} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{11\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)=\left(\cos\left(\frac{11\pi}{6}\right),\sin\left(\frac{11\pi}{6}\right)\right)$

$\cos\left(\frac{11\pi}{6} \right )=\frac{\sqrt{3}}{2}$ .

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Example Question #8 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\sin\left(\frac{11\pi}{6} \right )$

Possible Answers:

$\sin\left(\frac{11\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

$\sin\left(\frac{11\pi}{6} \right )=-\frac{1}{2}$

$\sin\left(\frac{11\pi}{6} \right )=\frac{1}{2}$

$\sin\left(\frac{11\pi}{6} \right )=-\frac{1}{\sqrt{2}}$

$\sin\left(\frac{11\pi}{6} \right )=\frac{\sqrt{3}}{2}$

Correct answer:

$\sin\left(\frac{11\pi}{6} \right )=-\frac{1}{2}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{11\pi}{6}$ .

$\theta = \frac{11\pi}{6}\Rightarrow \left(\frac{\sqrt{3}}{2}, -\frac{1}{2} \right )$

Step 3: Calculate the exact value of $\sin\left(\frac{11\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)=\left(\cos\left(\frac{11\pi}{6}\right),\sin\left(\frac{11\pi}{6}\right)\right)$

$\sin\left(\frac{11\pi}{6} \right )=-\frac{1}{2}$ .

Report an Error

Example Question #9 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\cos\left(\frac{\pi}{4} \right )$

Possible Answers:

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

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

$\cos\left(\frac{\pi}{4} \right )=\sqrt{2}$

$\cos\left(\frac{\pi}{4} \right )=-\frac{1}{\sqrt{2}}$

$\cos\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Correct answer:

$\cos\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{\pi}{4}$ .

$\theta = \frac{\pi}{4}\Rightarrow \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{\pi}{4} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\left(\cos\left(\frac{\pi}{4}\right),\sin\left(\frac{\pi}{4}\right)\right)$

$\cos\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$ .

Report an Error

Example Question #10 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

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

Possible Answers:

$\sin\left(\frac{\pi}{4} \right )=-\frac{1}{\sqrt{2}}$

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

$\sin\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

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

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

Correct answer:

$\sin\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{\pi}{4}$ .

$\theta = \frac{\pi}{4}\Rightarrow \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right )$

Step 3: Calculate the exact value of $\sin\left(\frac{\pi}{4} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\left(\cos\left(\frac{\pi}{4}\right),\sin\left(\frac{\pi}{4}\right)\right)$

$\sin\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}$ .

Report an Error

Example Question #41 : Trigonometric Functions

Find the following trigonometric exact value.

$\cos\left(\frac{3\pi}{4} \right )$

Possible Answers:

$\cos\left(\frac{3\pi}{4} \right )=\frac{1}{2}$

$\cos\left(\frac{3\pi}{4} \right )=-\sqrt{2}$

$\cos\left(\frac{3\pi}{4} \right )=-\frac{1}{2}$

$\cos\left(\frac{3\pi}{4} \right )=-\frac{1}{\sqrt{2}}$

$\cos\left(\frac{3\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Correct answer:

$\cos\left(\frac{3\pi}{4} \right )=-\frac{1}{\sqrt{2}}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{3\pi}{4}$ .

$\theta = \frac{3\pi}{4}\Rightarrow \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{3\pi}{4} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\left(\cos\left(\frac{3\pi}{4}\right),\sin\left(\frac{3\pi}{4}\right)\right)$

$\cos\left(\frac{3\pi}{4} \right )=-\frac{1}{\sqrt{2}}$ .

Report an Error

Example Question #42 : Trigonometric Functions

Find the following trigonometric exact value.

$\sin\left(\frac{3\pi}{4} \right )$

Possible Answers:

$\sin\left(\frac{3\pi}{4} \right )=\frac{1}{\sqrt{2}}$

$\sin\left(\frac{3\pi}{4} \right )=-\frac{1}{\sqrt{2}}$

$\sin\left(\frac{3\pi}{4} \right )=-\frac{1}{2}$

$\sin\left(\frac{3\pi}{4} \right )=\sqrt{2}$

$\sin\left(\frac{3\pi}{4} \right )=\frac{1}{2}$

Correct answer:

$\sin\left(\frac{3\pi}{4} \right )=\frac{1}{\sqrt{2}}$

Explanation:

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the (x,y) coordinate pair that represents the extension of $\theta = \frac{3\pi}{4}$ .

$\theta = \frac{3\pi}{4}\Rightarrow \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right )$

Step 3: Calculate the exact value of $\sin\left(\frac{3\pi}{4} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\left(\cos\left(\frac{3\pi}{4}\right),\sin\left(\frac{3\pi}{4}\right)\right)$

$\sin\left(\frac{3\pi}{4} \right )=\frac{1}{\sqrt{2}}$ .

Report an Error

Example Question #1 : Model Periodicity, Amplitude, Frequency, And Midline Of Trigonometric Functions: Ccss.Math.Content.Hsf Tf.B.5

What is the period of the following function?

$y=\sin(2x)$

Possible Answers:

$\frac{\pi}{2}$

$2\pi$

$\pi$

Correct answer:

$\pi$

Explanation:

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Write the general form of trigonometric shifts.

F(x)=af(bx-c)+d

where

$\\F(x)=\textup{New function after shift} \\a=\textup{Amplitude} \\b=\textup{period}\rightarrow \frac{2\pi}{|B|} \\ \\c=\textup{Horizontal shift} \\ d=\textup{Vertical shift} \\f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the period.

Given the function

$y=\sin(2x)$

identify the variables of the general form.

$\\f(x)=\sin \\a=1 \\b=2\rightarrow \frac{2\pi}{|2|}=\pi \\c=0 \\d=0$

Step 3: Graph the trigonometric function to verify.

Screen shot 2016 01 19 at 6.51.48 am

The graph above verifies that the period of this function is $\pi$ because the flow of the graph only begins to repeat its cycle after $\pi$ units on the - axis.

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Example Question #1 : Model Periodicity, Amplitude, Frequency, And Midline Of Trigonometric Functions: Ccss.Math.Content.Hsf Tf.B.5

What is the amplitude of the following function?

$y=2\sin(x)$

Possible Answers:

$\frac{\pi}{2}$

$2\pi$

Correct answer:

Explanation:

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Write the general form of trigonometric shifts.

F(x)=af(bx-c)+d

where

Step 2: Algebraically identify the amplitude.

Given the function

$y=2\sin(x)$

identify the variables of the general form.

$\\f(x)=\sin \\a=2 \\b=1\rightarrow \frac{2\pi}{|1|}=2\pi \\c=0 \\d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the amplitude of this function is two because the range of the function on the graph goes from negative two to positive two meaning the distance from zero at its highest peak or lowest valley is two.

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All Common Core: High School - Functions Resources

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Common Core: High School - Functions : Trigonometric Functions

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #2 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #1 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #1 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #8 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #9 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #10 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #41 : Trigonometric Functions

Example Question #42 : Trigonometric Functions

Example Question #1 : Model Periodicity, Amplitude, Frequency, And Midline Of Trigonometric Functions: Ccss.Math.Content.Hsf Tf.B.5

Example Question #1 : Model Periodicity, Amplitude, Frequency, And Midline Of Trigonometric Functions: Ccss.Math.Content.Hsf Tf.B.5

All Common Core: High School - Functions Resources

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