We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Common Core: High School - Functions : Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

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

Create An Account

All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 Next →

Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for sine and special reference angles calculate,

$\sin(105^\circ)$ .

Possible Answers:

$\sin(105^\circ)=\frac{\sqrt{2}}{2}$

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

$\sin(105^\circ)=\frac{\sqrt{8}}{2}$

$\sin(105^\circ)=\frac{2+\sqrt{6}}{2}$

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{2}$

Correct answer:

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

Explanation:

This type of question tests the deep understanding of geometry, right triangles, trigonometry, and dealing with proofs. Questions such as these are not designed to be tested but instead are used to build knowledge that will help in higher level mathematics courses.

For the purpose of Common Core Standards, "prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems", falls within the Cluster C of "prove and apply trigonometric identities" (CCSS.MATH.CONTENT.HSF.TF.C).

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: Break the angle into two angles that correspond to special reference angles.

$105^\circ=60^\circ+45^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=60^\circ \\b=45^\circ$

$\\ \sin(60^\circ+45^\circ)=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ) \\\sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{2}\times\frac{1}{\sqrt{2}}+\frac{1}{2}\times\frac{1}{\sqrt{2}} \\ \\ \sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ \\ \sin(60^\circ+45^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}$

To rationalize the denominator multiply the numerator and denominator by the square root of two.

$\\ \sin(60^\circ+45^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4} \\ \\ \sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

Report an Error

Example Question #2 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for sine and special reference angles calculate,

$\sin(75^\circ)$ .

Possible Answers:

$\sin(75^\circ)=\frac{1+\sqrt{3}}{\sqrt2}$

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2}$

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2\sqrt2}$

$\sin(75^\circ)=\frac{1}{2\sqrt2}$

$\sin(75^\circ)=\frac{\sqrt{3}}{2\sqrt2}$

Correct answer:

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2\sqrt2}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$75^\circ=30^\circ+45^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=30^\circ \\b=45^\circ$

$\\ \sin(30^\circ+45^\circ)=\sin(30^\circ)\cos(45^\circ)+\cos(30^\circ)\sin(45^\circ) \\\sin(30^\circ+45^\circ)=\frac{1}{2}\times\frac{1}{\sqrt{2}}+\frac{\sqrt3}{2}\times\frac{1}{\sqrt{2}} \\ \\ \sin(30^\circ+45^\circ)=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}} \\ \\ \sin(30^\circ+45^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}$

Report an Error

Example Question #3 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for sine and special reference angles calculate,

$\sin(120^\circ)$ .

Possible Answers:

$\sin(120^\circ)=-\frac{\sqrt{3}}{4}$

$\sin(120^\circ)=\frac{\sqrt{3}}{4}$

$\sin(120^\circ)=\frac{\sqrt{3}}{2\sqrt{2}}$

$\sin(120^\circ)=-\frac{\sqrt{3}}{2}$

$\sin(120^\circ)=\frac{\sqrt{3}}{2}$

Correct answer:

$\sin(120^\circ)=\frac{\sqrt{3}}{2}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$120^\circ=60^\circ+60^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=60^\circ \\b=60^\circ$

$\\ \sin(60^\circ+60^\circ)=\sin(60^\circ)\cos(60^\circ)+\cos(60^\circ)\sin(60^\circ) \\\sin(60^\circ+60^\circ)=\frac{\sqrt{3}}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2} \\ \\ \sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \\ \sin(60^\circ+60^\circ)=\frac{2\sqrt{3}}{4}$

$\\ \sin(60^\circ+60^\circ)=\frac{\sqrt{3}}{2}\\ \\ \sin(120^\circ)=\frac{\sqrt{3}}{2}$

Report an Error

Example Question #4 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for sine and special reference angles calculate,

$\sin(135^\circ)$ .

Possible Answers:

$\sin(135^\circ)=\frac{1}{2}$

$\sin(135^\circ)={\sqrt{2}}$

$\sin(135^\circ)=\frac{1}{\sqrt{2}}$

$\sin(135^\circ)=-\frac{1}{2}$

$\sin(135^\circ)=-\frac{1}{\sqrt{2}}$

Correct answer:

$\sin(135^\circ)=\frac{1}{\sqrt{2}}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$135^\circ=90^\circ+45^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=90^\circ \\b=45^\circ$

$\\ \sin(90^\circ+45^\circ)=\sin(90^\circ)\cos(45^\circ)+\cos(90^\circ)\sin(45^\circ) \\\sin(90^\circ+45^\circ)=1\times\frac{1}{\sqrt{2}}+0\times\frac{1}{\sqrt{2}} \\ \\ \sin(90^\circ+45^\circ)=\frac{1}{\sqrt{2}}$

Report an Error

Example Question #5 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for sine and special reference angles calculate,

$\sin(450^\circ)$ .

Possible Answers:

$\sin(450^\circ)=-1$

$\sin(450^\circ)=\frac{1}{2}$

$\sin(450^\circ)=1$

$\sin(450^\circ)=\frac{1}{2}+\sqrt{3}$

$\sin(450^\circ)=-\frac{1}{2}$

Correct answer:

$\sin(450^\circ)=1$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$450^\circ=360^\circ+90^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=360^\circ \\b=90^\circ$

$\sin(360+90)=\sin(360)\cos(90)+\cos(360)\sin(90)$

$\sin(360+90)=0\times 0+1\times1$

$\sin(360+90)=1$

$\sin(450^\circ)=1$

Report an Error

Example Question #6 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the subtraction formula for sine and special reference angles calculate,

$\sin(15^\circ)$ .

Possible Answers:

$\sin(15^\circ)= \frac{\sqrt6-1}{4}$

$\sin(15^\circ)= \frac{\sqrt6+\sqrt{2}}{4}$

$\sin(15^\circ)= \frac{\sqrt6+\sqrt{2}}{2}$

$\sin(15^\circ)= \frac{\sqrt6-\sqrt{2}}{4}$

$\sin(15^\circ)= \frac{\sqrt6-\sqrt{2}}{2}$

Correct answer:

$\sin(15^\circ)= \frac{\sqrt6-\sqrt{2}}{4}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$15^\circ=45^\circ-30^\circ$

Step 2: Write the general addition formula for sine.

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for sine.

$\\a=45^\circ \\b=30^\circ$

$\sin(45-30)=\sin(45)\cos(30)-\cos(45)\sin(30)$

$\sin(45-30)=\frac{1}{\sqrt{2}}\times \frac{\sqrt3}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$

$\sin(45-30)= \frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}$

$\sin(45-30)= \frac{\sqrt3-1}{2\sqrt{2}}$

Now to rationalize the denominator multiply the numerator and denominator by the square root of two.

$\sin(45-30)= \frac{\sqrt3-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$\sin(45-30)= \frac{\sqrt6-\sqrt{2}}{2\times 2}$

$\sin(45-30)= \frac{\sqrt6-\sqrt{2}}{4}$

$\sin(15^\circ)= \frac{\sqrt6-\sqrt{2}}{4}$

Report an Error

Example Question #7 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the subtraction formula for cosine and special reference angles calculate,

$\cos(15^\circ)$ .

Possible Answers:

$\cos(15^\circ)=\frac{\sqrt{6}+\sqrt{2}}{2}$

$\cos(15^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos(15^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}$

$\cos(15^\circ)=\frac{\sqrt{6}+1}{4}$

$\cos(15^\circ)=\frac{\sqrt{6}-\sqrt{2}}{2}$

Correct answer:

$\cos(15^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$15^\circ=45^\circ-30^\circ$

Step 2: Write the general addition formula for cosine.

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for cosine.

$\\a=45^\circ \\b=30^\circ$

$\cos(45-30)=\cos(45)\cos(30)+\sin(45)\sin(30)$

$\cos(45-30)=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}$

$\cos(45-30)=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}$

$\cos(45-30)=\frac{\sqrt{3}+1}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by the square root of two.

$\cos(45-30)=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos(15^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

Report an Error

Example Question #8 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for cosine and special reference angles calculate,

$\cos(105^\circ)$ .

Possible Answers:

$\cos(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

$\cos(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{2}$

$\cos(105^\circ)=\frac{1-\sqrt{6}}{4}$

$\cos(105^\circ)=\frac{\sqrt{2}-\sqrt{6}}{4}$

$\cos(105^\circ)=\frac{\sqrt{2}-\sqrt{6}}{2}$

Correct answer:

$\cos(105^\circ)=\frac{\sqrt{2}-\sqrt{6}}{4}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$105^\circ=60^\circ+45^\circ$

Step 2: Write the general addition formula for cosine.

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for cosine.

$\\a=60^\circ \\b=45^\circ$

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(60+45)=\cos(60)\cos(45)-\sin(60)\sin(45)$

$\cos(60+45)=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}$

$\cos(60+45)=\frac{1-\sqrt{3}}{2\sqrt{2}}$

To rationalize the denominator multiply the numerator and denominator by the square root of two.

$\cos(60+45)=\frac{1-\sqrt{3}}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$

$\cos(60+45)=\frac{\sqrt{2}-\sqrt{6}}{4}$

$\cos(105^\circ)=\frac{\sqrt{2}-\sqrt{6}}{4}$

Report an Error

Example Question #9 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for cosine and special reference angles calculate,

$\cos(450^\circ)$ .

Possible Answers:

$\cos(450^\circ)=-\frac{1}{2}$

$\cos(450^\circ)=-1$

$\cos(450^\circ)=\frac{1}{2}$

$\cos(450^\circ)=0$

$\cos(450^\circ)=1$

Correct answer:

$\cos(450^\circ)=0$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$450^\circ=360^\circ+90^\circ$

Step 2: Write the general addition formula for cosine.

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for cosine.

$\\a=360^\circ \\b=90^\circ$

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(360+90)=\cos(360)\cos(90)-\sin(360)\sin(90)$

$\cos(360+90)=1(0)-0(1)$

$\cos(360+90)=0$

$\cos(450^\circ)=0$

Report an Error

Example Question #10 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Using the addition formula for cosine and special reference angles calculate,

$\cos(75^\circ)$

Possible Answers:

$\cos(75^\circ)=\frac{\sqrt{6}-1}{4}$

$\cos(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{2}$

$\cos(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{2}$

$\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}$

Correct answer:

$\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}$

Explanation:

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: Break the angle into two angles that correspond to special reference angles.

$75^\circ=45^\circ+30^\circ$

Step 2: Write the general addition formula for cosine.

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for cosine.

$\\a=45^\circ \\b=30^\circ$

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(45+30)=\cos(45)\cos(30)-\sin(45)\sin(30)$

$\cos(45+30)=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2}$

$\cos(45+30)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

To rationalize the denominator, multiply by the square root of two.

$\cos(45+30)=\frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$\cos(45+30)=\frac{\sqrt{6}-\sqrt{2}}{4}$

$\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}$

Report an Error

← Previous 1 2 Next →

View Tutors

Brian
Certified Tutor

Georgia Institute of Technology-Main Campus, Bachelor of Science, Economics.

View Tutors

Mohammed
Certified Tutor

University of Kent, Bachelor in Arts, Prelaw Studies. Buckinghamshire New University, Masters in Business Administration, Bus...

View Tutors

Sagarika
Certified Tutor

All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in San Diego, Algebra Tutors in Denver, GRE Tutors in San Diego, Spanish Tutors in Denver, Math Tutors in Miami, SAT Tutors in Phoenix, English Tutors in Dallas Fort Worth, GMAT Tutors in Miami, Spanish Tutors in Los Angeles, Computer Science Tutors in Houston

Popular Courses & Classes

ACT Courses & Classes in Denver, ACT Courses & Classes in Miami, SSAT Courses & Classes in Seattle, SAT Courses & Classes in Houston, SAT Courses & Classes in Atlanta, SAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Houston, SAT Courses & Classes in Miami, SSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Washington DC

Popular Test Prep

GRE Test Prep in Atlanta, ACT Test Prep in Denver, ISEE Test Prep in Washington DC, GMAT Test Prep in Washington DC, MCAT Test Prep in Phoenix, GMAT Test Prep in Dallas Fort Worth, SSAT Test Prep in Philadelphia, SAT Test Prep in Chicago, GRE Test Prep in Phoenix, GRE Test Prep in Washington DC

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Common Core: High School - Functions : Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

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

All Common Core: High School - Functions Resources

Example Questions

Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #2 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #3 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #4 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #5 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #6 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #7 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #8 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #9 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #10 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

All Common Core: High School - Functions Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: