Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Sum and Difference Identities For Cosine

Study concepts, example questions & explanations for Precalculus

CREATE AN ACCOUNT Create Tests & Flashcards

Home Embed

All Precalculus Resources

12 Diagnostic Tests 400 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #13 : Trigonometric Identities

Evaluate the exact value of:

2cos(75)

Possible Answers:

$\frac{\sqrt6+\sqrt2}{2}$

$\frac{\sqrt6+\sqrt2}{4}$

$\frac{\sqrt6-\sqrt2}{2}$

$\frac{\sqrt6-\sqrt2}{4}$

$\frac{\sqrt2-\sqrt6}{4}$

Correct answer:

$\frac{\sqrt6-\sqrt2}{2}$

Explanation:

In order to solve 2cos(75) , two special angles will need to be used to solve for the exact values.

The angles chosen are A=30 and B=45 degrees, since:

A+B=30+45 =75

Write the formula for the cosine additive identity.

cos(A+B)= cosAcosB-sinAsinB

Substitute the known variables.

$cos(30)cos(45)-sin(30)sin(45)=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)$

$=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{4}$

$2cos(75)=2\left(\frac{\sqrt6-\sqrt2}{4}\right)=\frac{\sqrt6-\sqrt2}{2}$

Report an Error

Example Question #14 : Trigonometric Identities

Find the exact value of cos(15)-cos(75) .

Possible Answers:

$\frac{\sqrt2}{2}$

$\sqrt2$

$\sqrt6+\sqrt2$

$\frac{\sqrt6+\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

To solve cos(15)-cos(75) , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for cos(15) and cos(75) , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

A=45

B=30

Substitute the special angles in the formula.

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

Evaluate both conditions.

cos(75)=cos(45+30)=cos(45)cos(30)+sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

cos(15)=cos(45-30)=cos(45)cos(30)-sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)-\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Solve for cos(15)-cos(75) .

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\left[\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\right]$

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$= \frac{2\sqrt2}{4}$

$= \frac{\sqrt2}{2}$

Report an Error

Example Question #15 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\cos (\alpha + \beta)$ .

Possible Answers:

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {1- 6\sqrt{10}}{20}$

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {1+ 6\sqrt{10}}{20}$

Correct answer:

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y=1 and r=5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x=1 and r=4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ . So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the cosine sum formula, we then see:

$\\ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} - \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} - \sqrt {15}}{20}$ .

Report an Error

Example Question #16 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

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

Possible Answers:

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {1- 6\sqrt{10}}{20}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {1+ 6\sqrt{10}}{20}$

Correct answer:

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the cosine difference formula, we see:
$\\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}$

Report an Error

View Pre-Calculus Tutors

Julio
Certified Tutor

CUNY Hunter College, Bachelor in Arts, Statistics. CUNY Hunter College, Master of Arts, Applied Mathematics.

View Pre-Calculus Tutors

Patrick
Certified Tutor

Boston College, Bachelor of Science, Finance.

View Pre-Calculus Tutors

Alena
Certified Tutor

Mogilevsky State Pedagogical Institute, Bachelor in Arts, Mathematics. Mogilevsky State Pedagogical Institute, Masters, Mathe...

All Precalculus Resources

12 Diagnostic Tests 400 Practice Tests Question of the Day Flashcards Learn by Concept

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

Precalculus : Sum and Difference Identities For Cosine

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #13 : Trigonometric Identities

Example Question #14 : Trigonometric Identities

Example Question #15 : Trigonometric Identities

Example Question #16 : Trigonometric Identities

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

FIND THE BEST TUTORS

Email address:
Your name:

Email address:
Your name:
Feedback: