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Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Find the exact value of the following statement.

$f(x)=\cos^{-1}(1)$ $\displaystyle f(x)=\cos^{-1}(1)$

$\pi$ $\displaystyle \pi$

$2\pi$ $\displaystyle 2\pi$

$-\pi$ $\displaystyle -\pi$

$\frac{\pi}{3}$ $\displaystyle \frac{\pi}{3}$

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

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Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Find the exact value of the following statement.

$f(x)=\cos^{-1}(1)$ $\displaystyle f(x)=\cos^{-1}(1)$

$\pi$ $\displaystyle \pi$

$2\pi$ $\displaystyle 2\pi$

$-\pi$ $\displaystyle -\pi$

$\frac{\pi}{3}$ $\displaystyle \frac{\pi}{3}$

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

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Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Find the exact value of the following statement.

$f(x)=\cos^{-1}(1)$ $\displaystyle f(x)=\cos^{-1}(1)$

Your answer:

$\pi$ $\displaystyle \pi$

Your answer:

$-\pi$ $\displaystyle -\pi$

Your answer:

$\frac{\pi}{3}$ $\displaystyle \frac{\pi}{3}$

Your answer:

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

Correct answer:

$2\pi$ $\displaystyle 2\pi$

Explanation:

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

For the purpose of Common Core Standards, " Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed." concept (CCSS.MATH.CONTENT.HSF-TF.B.6). 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: Identify what the question is asking for.

Since there is a trigonometric function raised to the negative one power, this question is talking about the inverse of the function. In other words, which angle on the unit circle results in a cosine equalling one?

$y=\cos^{-1}(a)\Rightarrow \cos(\theta)=a$ $\displaystyle y=\cos^{-1}(a)\Rightarrow \cos(\theta)=a$

Therefore, theta needs to be solved for.

Step 2: Draw and label the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 3: Locate the angle that results in one for its cosine value.

Recall that

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

therefore look for the (x,y) $\displaystyle (x,y)$ that has x=1 $\displaystyle x=1$ . Looking at the unit circle from step 2, it is seen that at angle $2\pi$ $\displaystyle 2\pi$ the cosine equals one.

Thus,

$f(x)=\cos^{-1}(1)\Rightarrow f(x)=2\pi$ $\displaystyle f(x)=\cos^{-1}(1)\Rightarrow f(x)=2\pi$

To verify the solution simply find the cosine of the angle theta.

$\cos(2\pi)=1\rightarrow\cos^{-1}(1)=2\pi$ $\displaystyle \cos(2\pi)=1\rightarrow\cos^{-1}(1)=2\pi$

Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Find the exact value of the following statement.

$f(x)=\sin^{-1}(-1)$ $\displaystyle f(x)=\sin^{-1}(-1)$

$\sin^{-1}(-1)=\frac{3\pi}{4}$ $\displaystyle \sin^{-1}(-1)=\frac{3\pi}{4}$

$\sin^{-1}(-1)=\frac{3\pi}{2}$ $\displaystyle \sin^{-1}(-1)=\frac{3\pi}{2}$

$\sin^{-1}(-1)=\pi$ $\displaystyle \sin^{-1}(-1)=\pi$

$\sin^{-1}(-1)=2\pi$ $\displaystyle \sin^{-1}(-1)=2\pi$

$\sin^{-1}(-1)=\frac{\pi}{2}$ $\displaystyle \sin^{-1}(-1)=\frac{\pi}{2}$

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Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Find the exact value of the following statement.

$f(x)=\sin^{-1}(-1)$ $\displaystyle f(x)=\sin^{-1}(-1)$

Your answer:

$\sin^{-1}(-1)=\frac{3\pi}{4}$ $\displaystyle \sin^{-1}(-1)=\frac{3\pi}{4}$

Your answer:

$\sin^{-1}(-1)=\pi$ $\displaystyle \sin^{-1}(-1)=\pi$

Your answer:

$\sin^{-1}(-1)=2\pi$ $\displaystyle \sin^{-1}(-1)=2\pi$

Your answer:

$\sin^{-1}(-1)=\frac{\pi}{2}$ $\displaystyle \sin^{-1}(-1)=\frac{\pi}{2}$

Correct answer:

$\sin^{-1}(-1)=\frac{3\pi}{2}$ $\displaystyle \sin^{-1}(-1)=\frac{3\pi}{2}$

Explanation:

This question is testing ones ability to understand and identify inverses of trigonometric functions 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: Identify what the question is asking for.

$y=\sin^{-1}(a)\Rightarrow \sin(\theta)=a$ $\displaystyle y=\sin^{-1}(a)\Rightarrow \sin(\theta)=a$

Therefore, theta needs to be solved for.

Step 2: Draw and label the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 3: Locate the angle that results in negative one for its sine value.

Recall that

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

therefore look for the (x,y) $\displaystyle (x,y)$ that has y=-1 $\displaystyle y=-1$ . Looking at the unit circle from step 2, it is seen that at angle $\frac{3\pi}{2}$ $\displaystyle \frac{3\pi}{2}$ the sine equals negative one.

Thus,

$f(x)=\sin^{-1}(-1)\Rightarrow f(x)=\frac{3\pi}{2}$ $\displaystyle f(x)=\sin^{-1}(-1)\Rightarrow f(x)=\frac{3\pi}{2}$

To verify the solution simply find the sine of the angle theta.

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

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Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

All Common Core: High School - Functions Resources

Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

Flashcards: Restricting Domain of Trigonometric Functions to Allow for Construction of Inverse: CCSS.Math.Content.HSF-TF.B.6

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