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Common Core: High School - Functions : Symmetry and Periodicity of Trigonometric Functions: CCSS.Math.Content.HSF.TF.B.4

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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.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\theta = \frac{\pi}{6}$ .

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

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

Recall that,

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

therefore,

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

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

Report an Error

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

Find the following trigonometric exact value.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\theta = \frac{\pi}{6}$ .

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

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

Recall that,

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

therefore,

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

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

Report an Error

Example Question #3 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Find the following trigonometric exact value.

$\cos\left(\frac{5\pi}{6} \right )$

Possible Answers:

$\cos\left(\frac{5\pi}{6} \right )=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

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

$\cos\left(\frac{5\pi}{6} \right )=\frac{1}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{1}{2}$

Correct answer:

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

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 $\theta = \frac{5\pi}{6}$ .

$\theta = \frac{5\pi}{6}\Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{2} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{5\pi}{6} \right )$ .

Recall that,

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

therefore,

$\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)=\left(\cos\left(\frac{5\pi}{6}\right),\sin\left(\frac{5\pi}{6}\right)\right)$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$ .

Report an Error

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

Find the following trigonometric exact value.

$\sin\left(\frac{5\pi}{6} \right )$

Possible Answers:

$\sin\left(\frac{5\pi}{6} \right )=\frac{1}{\sqrt{2}}$

$\sin\left(\frac{5\pi}{6} \right )=-\frac{1}{\sqrt{2}}$

$\sin\left(\frac{5\pi}{6} \right )=\frac{1}{2}$

$\sin\left(\frac{5\pi}{6} \right )=-\frac{1}{2}$

$\sin\left(\frac{5\pi}{6} \right )=\frac{1}{\sqrt{3}}$

Correct answer:

$\sin\left(\frac{5\pi}{6} \right )=\frac{1}{2}$

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 $\theta = \frac{5\pi}{6}$ .

$\theta = \frac{5\pi}{6}\Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{2} \right )$

Step 3: Calculate the exact value of $\sin\left(\frac{5\pi}{6} \right )$ .

Recall that,

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

therefore,

$\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)=\left(\cos\left(\frac{5\pi}{6}\right),\sin\left(\frac{5\pi}{6}\right)\right)$

$\sin\left(\frac{5\pi}{6} \right )=\frac{1}{2}$ .

Report an Error

Example Question #5 : 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:

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

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

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

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

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

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.

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,

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

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

Report an Error

Example Question #6 : 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:

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

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

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

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

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

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 #7 : 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:

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

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

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

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

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

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,

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

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

Report an Error

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:

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

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

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

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

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

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:

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

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

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

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

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

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:

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

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

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

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

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

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

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Common Core: High School - Functions : Symmetry and Periodicity of Trigonometric Functions: CCSS.Math.Content.HSF.TF.B.4

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

All Common Core: High School - Functions Resources

Example Questions

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

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

Example Question #3 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

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

Example Question #5 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #6 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4

Example Question #7 : 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

All Common Core: High School - Functions Resources

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