Trigonometric Functions - Common Core: High School - Functions

Card 0 of 432

Question

Solve the following for $\theta$

$2\tan(\theta)+4=6$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$2\tan(\theta)+4=6$

First subtract four from each side.

$2\tan(\theta)+4{\color{Red} -4}=6{\color{Red} -4}$

$2\tan(\theta)=2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(1)$

$\theta=\tan^{-1}(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{4} \right )=\frac{\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{1}=1$

$\tan\left(\frac{5\pi}{4} \right )=\frac{-\frac{1}{\sqrt2}}{-\frac{1}{\sqrt2}}=-\frac{1}{\sqrt2}\cdot -\frac{\sqrt2}{1}=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

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

Compare your answer with the correct one above

Question

Solve the following for $\theta$ .

$2\tan(\theta)=2$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2$

Divide each side by two.

$\frac{2\tan(\theta)}{2}=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(1)$

$\theta=\tan^{-1}(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{4} \right )=\frac{\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{1}=1$

$\tan\left(\frac{5\pi}{4} \right )=\frac{-\frac{1}{\sqrt2}}{-\frac{1}{\sqrt2}}=-\frac{1}{\sqrt2}\cdot -\frac{\sqrt2}{1}=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{4},\frac{5\pi}{4}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$\tan(\theta)+4=\sqrt{3}+4$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$\tan(\theta)+4=\sqrt{3}+4$

First subtract four from both sides.

$\\tan(\theta)+4{\color{Red} -4}=\sqrt{3}+4{\color{Red} -4} \\tan(\theta)=\sqrt{3}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(\sqrt{3})$

$\theta=\tan^{-1}(\sqrt{3})$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{3} \right )=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\cdot \frac{2}{1}=\sqrt{3}$

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

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3},\frac{4\pi}{3}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$2\tan(\theta)=2\sqrt{3}$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2\sqrt{3}$

Divide by two on both sides.

$\frac{2\tan(\theta)}{2}=\frac{2\sqrt{3}}{2}$

$\tan(\theta)=\sqrt{3}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(\sqrt{3})$

$\theta=\tan^{-1}(\sqrt{3})$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{3} \right )=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\cdot \frac{2}{1}=\sqrt{3}$

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

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3},\frac{4\pi}{3}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$3\tan(\theta)=\frac{-3}{\sqrt{3}}$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{-3}{\sqrt{3}}$

Divide by three on both sides.

$\frac{3\tan(\theta)}{3}=\frac{-3}{3\sqrt{3}}$

$\tan(\theta)=-\frac{1}{\sqrt{3}}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

$\theta=\tan^{-1}\left(-\frac{1}{\sqrt{3}} \right )$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{5\pi}{6} \right )=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{2}\cdot -\frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

$\tan\left(\frac{11\pi}{6} \right )=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{5\pi}{6},\frac{11\pi}{6}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$3\tan(\theta)=\frac{3}{\sqrt{3}}$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{3}{\sqrt{3}}$

Divide by three on both sides.

$\frac{3\tan(\theta)}{3}=\frac{3}{3\sqrt{3}}$

$\tan(\theta)=\frac{1}{\sqrt{3}}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\theta=\tan^{-1}\left(\frac{1}{\sqrt{3}} \right )$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{6} \right )=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\tan\left(\frac{7\pi}{6} \right )=\frac{-\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=-\frac{1}{2}\cdot -\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

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

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$\tan(\theta)-1=\frac{1}{\sqrt{3}}-1$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$\tan(\theta)-1=\frac{1}{\sqrt{3}}-1$

Add one to both sides.

$\tan(\theta)-1{\color{Red} +1}=\frac{1}{\sqrt{3}}-1{\color{Red} +1}$

$\tan(\theta)=\frac{1}{\sqrt{3}}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\theta=\tan^{-1}\left(\frac{1}{\sqrt{3}} \right )$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{\pi}{6} \right )=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\tan\left(\frac{7\pi}{6} \right )=\frac{-\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=-\frac{1}{2}\cdot -\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

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

Compare your answer with the correct one above

Question

Solve the following for $\theta$ .

$2\tan(\theta)-4=-6$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$2\tan(\theta)-4=-6$

First add four from each side.

$2\tan(\theta)-4{\color{Red} +4}=-6{\color{Red} +4}$

$2\tan(\theta)=-2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}=\frac{-2}{2}$

$\tan(\theta)=-1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(-1)$

$\theta=\tan^{-1}(-1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{3\pi}{4} \right )=\frac{\frac{1}{\sqrt2}}{-\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}\cdot -\frac{\sqrt2}{1}=-1$

$\tan\left(\frac{7\pi}{4} \right )=\frac{-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=-\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{1}=-1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$\tan(\theta)+1=-\sqrt{3}+1$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$\tan(\theta)+1=-\sqrt{3}+1$

Subtract one from each side.

$\tan(\theta)+1{\color{Red} -1}=-\sqrt{3}+1{\color{Red} -1}$

$\tan(\theta)=-\sqrt{3}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(-\sqrt{3})$

$\theta=\tan^{-1}(-\sqrt{3})$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

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

$\tan\left(\frac{5\pi}{3} \right )=\frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{-\sqrt{3}}{2}\cdot \frac{2}{1}=-\sqrt{3}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{2\pi}{3},\frac{5\pi}{3}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$2\tan(\theta)=-2\sqrt{3}$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$2\tan(\theta)=-2\sqrt{3}$

Divide by two on each side.

$\frac{2\tan(\theta)}{2}=\frac{-2\sqrt{3}}{2}$

$\tan(\theta)=-\sqrt{3}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(-\sqrt{3})$

$\theta=\tan^{-1}(-\sqrt{3})$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

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

$\tan\left(\frac{5\pi}{3} \right )=\frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{-\sqrt{3}}{2}\cdot \frac{2}{1}=-\sqrt{3}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{2\pi}{3},\frac{5\pi}{3}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$\tan(\theta)+1=1-\frac{1}{\sqrt{3}}$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$\tan(\theta)+1=1-\frac{1}{\sqrt{3}}$

Subtract one from each side.

$\tan(\theta)+1{\color{Red} -1}=1-\frac{1}{\sqrt{3}}{\color{Red} -1}$

$\tan(\theta)=-\frac{1}{\sqrt{3}}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

$\theta=\tan^{-1}\left(-\frac{1}{\sqrt{3}} \right )$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{5\pi}{6} \right )=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{1}{2}\cdot -\frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

$\tan\left(\frac{11\pi}{6} \right )=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{5\pi}{6},\frac{11\pi}{6}$

Compare your answer with the correct one above

Question

Solve for $\theta$ .

$-\tan(\theta)+2=3$

Answer

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

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: Use algebraic operations to manipulate the function.

$-\tan(\theta)+2=3$

Subtract two from both sides.

$-\tan(\theta)+2{\color{Red} -2}=3{\color{Red} -2}$

$-\tan(\theta)=1$

Now divide by negative one on both sides.

$\tan(\theta)=-1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(-1)$

$\theta=\tan^{-1}(-1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{3\pi}{4} \right )=\frac{\frac{1}{\sqrt2}}{-\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}\cdot -\frac{\sqrt2}{1}=-1$

$\tan\left(\frac{7\pi}{4} \right )=\frac{-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=-\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{1}=-1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Compare your answer with the correct one above

Question

What is the period of the following function?

$y=\sin(2x)$

Answer

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.

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.

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.

Compare your answer with the correct one above

Question

What is the amplitude of the following function?

$y=2\sin(x)$

Answer

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.

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 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.

Compare your answer with the correct one above

Question

What is the amplitude of the following function?

$y=4\sin(x)$

Answer

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.

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 amplitude.

Given the function

$y=4\sin(x)$

identify the variables of the general form.

$\f(x)=\sin \a=4 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

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

Compare your answer with the correct one above

Question

What is the period of the following function?

$y=\sin(4x)$

Answer

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.

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

$y=\sin(4x)$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=4\rightarrow \frac{2\pi}{|4|}=\frac{\pi}{2} \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the period of this function is $\frac{\pi}{2}$ because the flow of the graph only begins to repeat its cycle after $\frac{\pi}{2}$ units on the - axis.

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Question

What is the vertical shift of the function?

$f(x)=\sin(x)+\pi$

Answer

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.

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 vertical shift.

Given the function

$y=\sin(x)+\pi$

identify the variables of the general form.

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

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $\pi$ .

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Question

What is the vertical shift of the function?

$f(x)=\sin(x)-\pi$

Answer

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.

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 vertical shift.

Given the function

$y=\sin(x)-\pi$

identify the variables of the general form.

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

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $-\pi$ .

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Question

What is the horizontal shift of the function?

$f(x)=\sin(x-\pi)$

Answer

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.

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 horizontal shift.

Given the function

$y=\sin(x-\pi)$

identify the variables of the general form.

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

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the horizontal shift of the function is $\pi$ .

Compare your answer with the correct one above

Question

What is the horizontal shift of the function?

$f(x)=\sin(x+\pi)$

Answer

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.

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 horizontal shift.

Given the function

$y=\sin(x+\pi)$

identify the variables of the general form.

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

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the horizontal shift of the function is $-\pi$ .

Compare your answer with the correct one above

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