Common Core: High School - Functions : High School: Functions

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

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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 #2 : 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.

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

Possible Answers:

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

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

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

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

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

Correct answer:

\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.

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 sine equalling negative one?

\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 

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

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

Thus,

\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.

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

Example Question #3 : 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.

\displaystyle f(x)=\cos^{-1}(0)

Possible Answers:

\displaystyle \cos^{-1}(0)=\frac{\pi}{3} \text{ or } \frac{3\pi}{2}

\displaystyle \cos^{-1}(0)=\frac{\pi}{4} \text{ or } \frac{3\pi}{4}

\displaystyle \cos^{-1}(0)={\pi}\text{ or } \frac{3\pi}{2}

\displaystyle \cos^{-1}(0)=\frac{\pi}{2} \text{ or } \frac{3\pi}{2}

\displaystyle \cos^{-1}(0)=\frac{\pi}{2} \text{ or } \frac{3\pi}{4}

Correct answer:

\displaystyle \cos^{-1}(0)=\frac{\pi}{2} \text{ or } \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.

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 zero?

\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 zero for its cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=0. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2} the cosine equals zero. 

Thus,

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

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

\displaystyle \\\cos\left(\frac{3\pi}{2} \right )=0\rightarrow\cos^{-1}(0)=\frac{3\pi}{2} \\ \\\cos\left(\frac{\pi}{2} \right )=0\rightarrow\cos^{-1}(0)=\frac{\pi}{2}

Example Question #4 : 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.

\displaystyle f(x)=\cos^{-1}\left(\frac{1}{2} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{5\pi}{3}

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{5\pi}{2}

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{2\pi}{3}

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2} \text{ or } \frac{5\pi}{3}

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2} \text{ or } \frac{5\pi}{2}

Correct answer:

\displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{5\pi}{3}

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 half?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=\frac{1}{2}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{\pi}{3} and \displaystyle \frac{5\pi}{3} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(\frac{1}{2} \right )\Rightarrow f(x)=\frac{5\pi}{3},\frac{\pi}{3}

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

\displaystyle \\\cos\left(\frac{5\pi}{3} \right )=\frac{1}{2}\rightarrow\cos^{-1}\frac{1}{2}=\frac{5\pi}{3} \\ \\\cos\left(\frac{\pi}{3} \right )=\frac{1}{2}\rightarrow\cos^{-1}\frac{1}{2}=\frac{\pi}{3}

Example Question #1 : 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.

\displaystyle f(x)=\cos^{-1}\left(-\frac{1}{2} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{4\pi}{3}

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{5\pi}{3}

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{3} \text{ or } \frac{5\pi}{2}

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{2} \text{ or } \frac{5\pi}{3}

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \text{ or } \frac{4\pi}{3}

Correct answer:

\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \text{ or } \frac{4\pi}{3}

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 negative one half?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=-\frac{1}{2}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(-\frac{1}{2} \right )\Rightarrow f(x)=\frac{4\pi}{3},\frac{2\pi}{3}

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

\displaystyle \\\cos\left(\frac{4\pi}{3} \right )=-\frac{1}{2}\rightarrow\cos^{-1}\left( -\frac{1}{2}\right)=\frac{5\pi}{3} \\ \\\cos\left(\frac{2\pi}{3} \right )=-\frac{1}{2}\rightarrow\cos^{-1}\left( -\frac{1}{2}\right)=\frac{2\pi}{3}

Example Question #6 : 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.

\displaystyle f(x)=\cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{3\pi}{4}

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{2}\text{ or }\frac{5\pi}{2}

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{2}\text{ or }\frac{5\pi}{4}

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{4}\text{ or }\frac{5\pi}{2}

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{4}\text{ or }\frac{5\pi}{4}

Correct answer:

\displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{4}\text{ or }\frac{5\pi}{4}

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 negative one over the square root of two?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=-\frac{1}{\sqrt{2}}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{3\pi}{4} and \displaystyle \frac{5\pi}{4} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(-\frac{1}{\sqrt{2}} \right )\Rightarrow f(x)=\frac{3\pi}{4},\frac{5\pi}{4}

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

\displaystyle \\\cos\left(\frac{3\pi}{4} \right )=-\frac{1}{\sqrt{2}}\rightarrow\cos^{-1}\left( -\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4} \\ \\\cos\left(\frac{5\pi}{4} \right )=-\frac{1}{\sqrt{2}}\rightarrow\cos^{-1}\left( -\frac{1}{\sqrt{2}}\right)=\frac{5\pi}{4}

Example Question #7 : 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.

\displaystyle f(x)=\cos^{-1}\left(\frac{1}{\sqrt{2}} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{5\pi}{4}

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{5\pi}{2}

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{7\pi}{4}

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{3\pi}{4}

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{3\pi}{4}\text{ or }\frac{7\pi}{4}

Correct answer:

\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}} \right )=\frac{\pi}{4}\text{ or }\frac{7\pi}{4}

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 over the square root of two?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=\frac{1}{\sqrt{2}}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{\pi}{4} and \displaystyle \frac{7\pi}{4} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(\frac{1}{\sqrt{2}} \right )\Rightarrow f(x)=\frac{\pi}{4},\frac{7\pi}{4}

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

\displaystyle \\\cos\left(\frac{\pi}{4} \right )=\frac{1}{\sqrt{2}}\rightarrow\cos^{-1}\left( \frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} \\ \\\cos\left(\frac{7\pi}{4} \right )=\frac{1}{\sqrt{2}}\rightarrow\cos^{-1}\left( \frac{1}{\sqrt{2}}\right)=\frac{7\pi}{4}

Example Question #8 : 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.

\displaystyle f(x)=\cos^{-1}\left(\frac{\sqrt{3}}{2} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{\pi}{3}\text{ or }\frac{5\pi}{3}

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{4\pi}{6}\text{ or }\frac{11\pi}{6}

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{\pi}{2}\text{ or }\frac{3\pi}{2}

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{\pi}{6}\text{ or }\frac{11\pi}{6}

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{\pi}{3}\text{ or }\frac{11\pi}{3}

Correct answer:

\displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2} \right )=\frac{\pi}{6}\text{ or }\frac{11\pi}{6}

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 the square root of three over two?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=\frac{\sqrt{3}}{2}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{\pi}{6} and \displaystyle \frac{11\pi}{6} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(\frac{\sqrt{3}}{2} \right )\Rightarrow f(x)=\frac{\pi}{6},\frac{11\pi}{6}

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

\displaystyle \\\cos\left(\frac{\pi}{6} \right )=\frac{\sqrt{3}}{2}\rightarrow\cos^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{6} \\ \\\cos\left(\frac{11\pi}{6} \right )=\frac{\sqrt{3}}{2}\rightarrow\cos^{-1}\frac{\sqrt{3}}{2}=\frac{11\pi}{6}

Example Question #9 : 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.

\displaystyle f(x)=\cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )

Possible Answers:

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{\pi}{6}\text{ or }\frac{7\pi}{6}

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{4\pi}{6}\text{ or }\frac{3\pi}{6}

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{\pi}{6}\text{ or }\frac{11\pi}{6}

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{5\pi}{6}\text{ or }\frac{7\pi}{6}

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{5\pi}{6}\text{ or }\frac{\pi}{6}

Correct answer:

\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )=\frac{5\pi}{6}\text{ or }\frac{7\pi}{6}

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 negative square root of three over two?

\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 the given cosine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle x=-\frac{\sqrt{3}}{2}. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \frac{5\pi}{6} and \displaystyle \frac{7\pi}{6} the cosine equals the given value. 

Thus,

\displaystyle f(x)=\cos^{-1}\left(-\frac{\sqrt{3}}{2} \right )\Rightarrow f(x)=\frac{5\pi}{6},\frac{7\pi}{6}

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

\displaystyle \\\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}\rightarrow\cos^{-1}\frac{-\sqrt{3}}{2}=\frac{5\pi}{6} \\ \\\cos\left(\frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}\rightarrow\cos^{-1}\frac{-\sqrt{3}}{2}=\frac{7\pi}{6}

Example Question #10 : 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.

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

Possible Answers:

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

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

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

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

\displaystyle \sin^{-1}(1)=\frac{\pi}{6}

Correct answer:

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

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 sine equalling one?

\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 the given sine value.

Recall that 

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

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

Thus,

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

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

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

Example Question #71 : Trigonometric Functions

Find the exact value of the following statement.

\displaystyle f(x)=\sin^{-1}(0)

Possible Answers:

\displaystyle \sin^{-1}(0)=\frac{2\pi}{5}

\displaystyle \sin^{-1}(0)=\pi \text{ or }2\pi

\displaystyle \sin^{-1}(0)=\frac{3\pi}{2}

\displaystyle \sin^{-1}(0)=\frac{\pi}{4}

\displaystyle \sin^{-1}(0)=\frac{\pi}{2}

Correct answer:

\displaystyle \sin^{-1}(0)=\pi \text{ or }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 sine equalling zero?

\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 the given sine value.

Recall that 

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

therefore look for the \displaystyle (x,y) that has \displaystyle y=0. Looking at the unit circle from step 2, it is seen that at angle \displaystyle \pi and \displaystyle 2\pi the sine equals the given value. 

Thus,

\displaystyle f(x)=\sin^{-1}(0)\Rightarrow f(x)=\pi \text{ or }2\pi

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

\displaystyle \\\sin\left({\pi}\right)=0\rightarrow\sin^{-1}(0)=\pi\\ \sin(2\pi)=0\rightarrow \sin^{-1}(0)=2\pi

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