Call Now to Set Up Tutoring:

(888) 888-0446

Trigonometry : Sum and Difference of Sines and Cosines

Study concepts, example questions & explanations for Trigonometry

Create An Account

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #631 : Trigonometry

What is the correct formula for the sum of two sines: sin(x) + sin(y) ?

Possible Answers:

$sin(x) + sin(y) = 2sin\frac{x + y}{2}cos\frac{x-y}{2}$

$sin(x)+sin(y) = 2sin\frac{x+y}{2}sin\frac{x-y}{2}$

$sin(x)+sin(y) = 2sin\frac{x+y}{2}cos\frac{x+y}{2}$

$sin(x)+sin(y) = 2cos\frac{x+y}{2}sin\frac{x-y}{2}$

Correct answer:

$sin(x) + sin(y) = 2sin\frac{x + y}{2}cos\frac{x-y}{2}$

Explanation:

This is a known trigonometry identity. Whenever you are adding two sine functions, you can plug and into the formula to solve for this sum

Report an Error

Example Question #2 : Sum And Difference Of Sines And Cosines

Solve for the following given that $x = 2\pi$ . Use the formula for the sum of two sines.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x)$

Possible Answers:

$\frac{1}{2}$

$\sqrt{3}$

$\pi$

Correct answer:

$\sqrt{3}$

Explanation:

We begin by considering our formula for the sum of two sines

$sin(x) + sin(y) = 2sin\frac{x + y}{2}cos\frac{x-y}{2}$

We will let $x = \frac{\pi}{3} + x$ and $y = \frac{\pi}{3} - x$ and plug these values into our formula.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x) = 2sin\frac{(\frac{\pi}{3} + x) + (\frac{\pi}{3} - x)}{2}cos\frac{(\frac{\pi}{3} + x) - (\frac{\pi}{3} - x)}{2}$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\frac{2\pi}{3}}{2}cos\frac{2x}{2}$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\pi}{3}cosx$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2(\frac{\sqrt{3}}{2})(1)$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = \sqrt{3}$

Report an Error

Example Question #3 : Sum And Difference Of Sines And Cosines

Solve for the following using the formula for the differences of two cosines. Do not simplify.

cos(2 + x) - cos(2 - x)

Possible Answers:

$\pi$

-2sin(2)sin(x)

-2cos(2)cos(x)

Correct answer:

-2sin(2)sin(x)

Explanation:

We begin by considering the formula for the differences of two cosines.

$cos(x) - cos(y) = -2sin\frac{x + y}{2}sin\frac{x - y}{2}$

We will let x = 2+x and y = 2 - x . Proceed by plugging these values into the formula.

$cos(2+x) - cos(2 -x) = -2sin\frac{(2+x) + (2 - x)}{2}sin\frac{(2+x) - (2-x)}{2}$

$cos(2+x) - cos(2 -x) = -2sin\frac{4}{2}sin\frac{2x}{2}$

cos(2+x) - cos(2 -x) = -2sin(2)sin(x)

Report an Error

Example Question #1 : Sum And Difference Of Sines And Cosines

Which of the following completes the identity $cos(\alpha + \beta) =$

Possible Answers:

$sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)$

$cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$sin(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

$cos(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

Correct answer:

$cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

Explanation:

This is a known trigonometry identity and has been proven to be true. It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

Report an Error

Example Question #5 : Sum And Difference Of Sines And Cosines

Solve for the following using the correct identity:

$sin(\frac{\pi}{4} + \frac{\pi}{6})$

Possible Answers:

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{2}}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\sqrt{6} + 2\sqrt{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6} + 2\sqrt{2}}{4}$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6}}{4}$

Correct answer:

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6} + 2\sqrt{2}}{4}$

Explanation:

To solve this problem we must use the identity

$sin(\alpha + \beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)$ . We will let $\alpha = \frac{\pi}{4}$ and $\beta = \frac{\pi}{6}$ .

$sin(\frac{\pi}{4} + \frac{\pi}{6}) = sin(\frac{\pi}{4})cos(\frac{\pi}{6}) + cos(\frac{\pi}{4})sin(\frac{\pi}{6})$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6} + 2\sqrt{2}}{4}$

Report an Error

Example Question #1 : Sum And Difference Of Sines And Cosines

True or False: To solve for a problem in the form of $cos(\alpha - \beta)$ , I use the identity $cos(\alpha) - cos(\beta) = -2sin\frac{\alpha + \beta}{2}sin\frac{\alpha - \beta}{2}$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

This answer is false. $cos(\alpha - \beta)$ is not the same as $cos(\alpha) - cos(\beta)$ .

For example, say $\alpha = \frac{2\pi}{3}$ and $\beta = \frac {\pi}{3}$

$cos(\frac{2\pi}{3} - \frac{\pi}{3}) = cos(\frac{\pi}{3}) = \frac{1}{2}$

$cos(\frac{2\pi}{3}) - cos(\frac{\pi}{3} = -\frac{1}{2} - \frac{1}{2} = -1$

And so the correct identity to use for this is

$cos(\alpha - \beta) = cos(\alpha)cos(\beta) +sin(\alpha)sin(\beta)$

Report an Error

Example Question #7 : Sum And Difference Of Sines And Cosines

Solve for the following using the correct identity:

cos(45 + 30)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The correct identity to use for this kind of problem is

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

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

$cos(45 + 30) = cos(\frac{\pi}{4})cos(\frac{\pi}{6}) - sin(\frac{\pi}{4})sin(\frac{\pi}{6})$

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

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

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

Report an Error

Example Question #2 : Sum And Difference Of Sines And Cosines

Which of the following is the correct to complete the following identity: $sin(\alpha - \beta) =$ ___?

Possible Answers:

$sin(\alpha)cos(\alpha) - cos(\beta)sin(\beta)$

$cos(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

$sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)$

$sin(\alpha)cos(\beta) - cos(\alpha)sin(\beta)$

Correct answer:

$sin(\alpha)cos(\beta) - cos(\alpha)sin(\beta)$

Explanation:

Report an Error

View Trigonometry Tutors

Ishaan
Certified Tutor

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

View Trigonometry Tutors

Venkata Naga Sreelalitapriya
Certified Tutor

University of Central Florida, Bachelor of Science, Biotechnology.

View Trigonometry Tutors

Selina
Certified Tutor

Rutgers University-New Brunswick, Current Undergrad Student, Health Services Administration.

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Trigonometry Tutors in Top Cities:

Popular Courses & Classes

GRE Courses & Classes in Denver, SSAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Phoenix, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in New York City, MCAT Courses & Classes in Seattle, SAT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in San Diego, ISEE Courses & Classes in Philadelphia, SAT Courses & Classes in New York City

Popular Test Prep

LSAT Test Prep in San Diego, GMAT Test Prep in San Diego, MCAT Test Prep in Phoenix, ISEE Test Prep in Boston, GRE Test Prep in Chicago, ISEE Test Prep in Dallas Fort Worth, GRE Test Prep in Phoenix, ACT Test Prep in Los Angeles, ACT Test Prep in New York City, GRE Test Prep in Boston

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

Trigonometry : Sum and Difference of Sines and Cosines

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #631 : Trigonometry

Example Question #2 : Sum And Difference Of Sines And Cosines

Example Question #3 : Sum And Difference Of Sines And Cosines

Example Question #1 : Sum And Difference Of Sines And Cosines

Example Question #5 : Sum And Difference Of Sines And Cosines

Example Question #1 : Sum And Difference Of Sines And Cosines

Example Question #7 : Sum And Difference Of Sines And Cosines

Example Question #2 : Sum And Difference Of Sines And Cosines

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: