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Common Core: High School - Functions : Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

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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 #432 : High School: Functions

Solve the following for $\theta$

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

Possible Answers:

$\theta=\frac{2\pi}{4},\frac{7\pi}{4}$

$\theta=\frac{\pi}{4},\frac{2\pi}{4}$

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

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

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

Correct answer:

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

Explanation:

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve the following for $\theta$ .

$2\tan(\theta)=2$

Possible Answers:

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

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

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

$\theta=\frac{\pi}{4},\frac{6\pi}{4}$

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

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #434 : High School: Functions

Solve for $\theta$ .

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

Possible Answers:

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

$\theta=\frac{6\pi}{3},\frac{4\pi}{3}$

$\theta=\frac{\pi}{2},\frac{4\pi}{2}$

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

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

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #435 : High School: Functions

Solve for $\theta$ .

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

Possible Answers:

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

$\theta=\frac{\pi}{6},\frac{5\pi}{3}$

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

$\theta=\frac{\pi}{2},\frac{4\pi}{3}$

$\theta=\frac{\pi}{3},\frac{4\pi}{2}$

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

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

Possible Answers:

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

$\theta=\frac{\pi}{6},\frac{7\pi}{3}$

$\theta=\frac{\pi}{6},\frac{4\pi}{6}$

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

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

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #437 : High School: Functions

Solve for $\theta$ .

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

Possible Answers:

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

$\theta=\frac{\pi}{6},\frac{4\pi}{6}$

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

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

$\theta=\frac{\pi}{6},\frac{7\pi}{3}$

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #438 : High School: Functions

Solve for $\theta$ .

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

Possible Answers:

$\theta=\frac{\pi}{6},\frac{7\pi}{3}$

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

$\theta=\frac{2\pi}{6},\frac{7\pi}{6}$

$\theta=\frac{\pi}{6},\frac{3\pi}{6}$

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

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #439 : High School: Functions

Solve the following for $\theta$ .

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

Possible Answers:

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

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

$\theta=\frac{3\pi}{4},\frac{9\pi}{4}$

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

$\theta=\frac{3\pi}{2},\frac{7\pi}{2}$

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

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

Possible Answers:

$\theta=\frac{2\pi}{3},\frac{5\pi}{6}$

$\theta=\frac{2\pi}{6},\frac{5\pi}{6}$

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

$\theta=\frac{2\pi}{6},\frac{5\pi}{3}$

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

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

Example Question #3 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

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

Possible Answers:

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

$\theta=\frac{2\pi}{3},\frac{5\pi}{6}$

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

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

$\theta=\frac{2\pi}{6},\frac{5\pi}{6}$

Correct answer:

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

Explanation:

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

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.

Screen shot 2016 01 14 at 10.52.42 am

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,

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

Report an Error

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Common Core: High School - Functions : Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

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

All Common Core: High School - Functions Resources

Example Questions

Example Question #432 : High School: Functions

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #434 : High School: Functions

Example Question #435 : High School: Functions

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #437 : High School: Functions

Example Question #438 : High School: Functions

Example Question #439 : High School: Functions

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #3 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

All Common Core: High School - Functions Resources

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