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All Common Core: High School - Functions Resources

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Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Solve the following for $\theta$

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

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

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

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

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

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

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Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Solve the following for $\theta$

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

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

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,

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

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Solve the following for $\theta$ .

$2\tan(\theta)=2$

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

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

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

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

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

Report an Error Copyright Notice

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Solve the following for $\theta$ .

$2\tan(\theta)=2$

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

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,

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

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All Common Core: High School - Functions Resources

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

Flashcards: Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7

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