Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Sum and Difference Identities

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : 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:

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

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

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

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

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

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 #23 : 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:

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

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

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

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

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

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

Example Question #24 : Trigonometric Identities

Find $\sin\bigg(\frac{\pi }{3} + \frac{\pi }{6}\bigg)$ using the sum identity.

Possible Answers:

1.5

$\frac{\pi }{2}$

$\frac{\sqrt{3}}{2}$

Correct answer:

Explanation:

Using the sum formula for sine,

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

where,

$a=\frac{\pi}{3}$ , $b=\frac{\pi}{6}$

yeilds:

$\sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)$

$=\frac{\sqrt{3}}2{}\cdot\frac{\sqrt{3}}2{}+\frac{1}{2}\cdot\frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}=1$ .

Report an Error

Example Question #25 : Trigonometric Identities

Calculate $\sin(75)$ .

Possible Answers:

$\frac{6+\sqrt{2}}{4}$

$\frac{1}2{}$

$\frac{\sqrt{2}+\sqrt{6}}{4}$

Correct answer:

$\frac{\sqrt{2}+\sqrt{6}}{4}$

Explanation:

Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

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

where

a=30 , b=45 .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

$=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}2{}$

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

Report an Error

Example Question #26 : Trigonometric Identities

Find the exact value for: $4\sin (15^{\circ})$

Possible Answers:

$\sqrt2-\frac{\sqrt2}{3}$

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

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

$\sqrt6-\sqrt2$

Correct answer:

$\sqrt6-\sqrt2$

Explanation:

In order to solve this question, it is necessary to know the sine difference identity.

sin(A-B)=sinAcosB-cosAsinB

The values of and must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

A=45

B=30

Substitute the values into the formula and solve.

sin(45-30)=(sin45)(cos30)-(cos45)(sin30)

$sin(15)=(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})-(\frac{\sqrt2}{2})(\frac{1}{2})= \frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Evaluate 4sin(15) .

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

Report an Error

Example Question #27 : Trigonometric Identities

Find the exact value of: $\sin(105^{\circ})$

Possible Answers:

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

$\frac{\sqrt2+\sqrt5}{4}$

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

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

$\frac{\sqrt2+3\sqrt2}{4}$

Correct answer:

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

Explanation:

In order to find the exact value of sin(105) , the sum identity of sine must be used. Write the formula.

sin(A+B)=sinAcosB+cosAsinB

The only possibilites of and are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

sin(45+60)=sin45cos60+cos45sin60

$sin(105)=(\frac{\sqrt2}{2})(\frac{1}{2})+(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})=\frac{\sqrt2+\sqrt6}{4}$

Report an Error

Example Question #28 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 sine sum formula, we see:

$\\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}$

Report an Error

Example Question #29 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 sine difference formula, we see:

$\\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}$

Report an Error

Example Question #31 : Trigonometric Identities

Evaluate

$\sin \left(\frac{5 \pi}{12}\right)$ .

Possible Answers:

$\frac{-1 - \sqrt 3 }{\sqrt 2 }$

$\frac{1}{\sqrt 2 }$

$\sqrt{\frac{3}{2}}$

$\frac{1+\sqrt3}{2 \sqrt 2 }$

$\frac{-1+\sqrt3}{2 \sqrt 2 }$

Correct answer:

$\frac{1+\sqrt3}{2 \sqrt 2 }$

Explanation:

$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 } + \frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 } + \frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

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

From the unit circle, we can determine these measures:

$\frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

Report an Error

Example Question #32 : Trigonometric Identities

Evaluate

$\sin(285 ^ o)$ .

Possible Answers:

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

$\frac { 1 - \sqrt 3 }{ 2 \sqrt 2 }$

$\frac{1+ \sqrt 3 }{ 2 \sqrt 2 }$

$\frac{\sqrt 3 - 1 }{4 \sqrt 2 }$

$\frac{\sqrt 3 - 1 }{2 \sqrt 2 }$

Correct answer:

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

Explanation:

The angle 285 ^o = 225^ o + 60 ^ o or 45^o + 240 ^ o .

Using the first one:

$\sin(60^o + 225^o ) = \sin (60^o) \cos (225^o) + \cos (60^o) \cos(225^o )$

We can find these values in the unit circle:

$\frac{\sqrt 3 }{2 } \cdot \frac{-1}{\sqrt 2 } + \frac{1}{2} \cdot \frac{-1}{\sqrt 2 } = \frac{-\sqrt3 }{2 \sqrt 2 } + \frac{-1}{2\sqrt2} = \frac{-\sqrt 3 - 1 }{2 \sqrt 2 }$

Report an Error

View Pre-Calculus Tutors

Tamera
Certified Tutor

Christopher Newport University, Bachelor of Science, Biochemistry.

View Pre-Calculus Tutors

James
Certified Tutor

Purdue University-Main Campus, Doctor of Education, Mathematics Teacher Education.

View Pre-Calculus Tutors

Jody
Certified Tutor

University of Kansas, Bachelor of Science, Mathematics Teacher Education. Maryville University of Saint Louis, Master of Scie...

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Chicago, Reading Tutors in Dallas Fort Worth, French Tutors in New York City, Spanish Tutors in New York City, GMAT Tutors in Atlanta, Physics Tutors in Denver, MCAT Tutors in Dallas Fort Worth, ISEE Tutors in Miami, ISEE Tutors in Denver, MCAT Tutors in Boston

Popular Courses & Classes

MCAT Courses & Classes in Washington DC, GMAT Courses & Classes in Atlanta, SAT Courses & Classes in Washington DC, SAT Courses & Classes in Atlanta, SSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Philadelphia, MCAT Courses & Classes in San Diego, SAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Atlanta

Popular Test Prep

GRE Test Prep in Washington DC, GRE Test Prep in San Diego, GMAT Test Prep in Philadelphia, SAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Los Angeles, GRE Test Prep in Philadelphia, MCAT Test Prep in Miami, SAT Test Prep in Denver, SAT Test Prep in San Francisco-Bay Area, SAT 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

Precalculus : Sum and Difference Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #22 : Trigonometric Identities

Example Question #23 : Trigonometric Identities

Example Question #24 : Trigonometric Identities

Example Question #25 : Trigonometric Identities

Example Question #26 : Trigonometric Identities

Example Question #27 : Trigonometric Identities

Example Question #28 : Trigonometric Identities

Example Question #29 : Trigonometric Identities

Example Question #31 : Trigonometric Identities

Example Question #32 : Trigonometric Identities

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: