Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Trigonometric 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 #1471 : Pre Calculus

Evaluate

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

Possible Answers:

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

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

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

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

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

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 #1472 : Pre Calculus

Evaluate

$\sin(285 ^ o)$ .

Possible Answers:

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

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

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

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

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

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

Example Question #31 : 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

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

Possible Answers:

$\frac {25\sqrt{6}-32\sqrt{15}}{9}$

$\frac {4}{5}$

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

$\frac {25\sqrt{6}+32\sqrt{15}}{9}$

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

Correct answer:

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

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
$\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

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 $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\\ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ \\ \\* = \frac{\frac{\sqrt{6}}{12}+\sqrt{15}}{1-\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}+12\sqrt{15}}{12}}{\frac{12-3\sqrt{10}}{12}} \\ \\ \\* = \frac{\sqrt{6}+12\sqrt{15}}{12-3\sqrt{10}} \cdot \frac{12+3\sqrt{10}}{12+3\sqrt{10}} \\ \\ \\* = \frac{12\sqrt{6}+6\sqrt{15}+144\sqrt{15}+180\sqrt{6}}{144-90} \\ \\* = \frac{192\sqrt{6}+150\sqrt{15}}{54} = \frac{32\sqrt{6}+25\sqrt{15}}{9}$

Report an Error

Example Question #1475 : Pre Calculus

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

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

Possible Answers:

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

$\frac {25\sqrt{6}+32\sqrt{15}}{9}$

$\frac {25\sqrt{6}-32\sqrt{15}}{9}$

$\frac{1}{9}$

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

Correct answer:

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

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 $\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

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 $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\\ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} \\ \\ \\* = \frac{\frac{\sqrt{6}}{12}-\sqrt{15}}{1+\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}-12\sqrt{15}}{12}}{\frac{12+3\sqrt{10}}{12}} \\ \\ \\* = \frac{\sqrt{6}-12\sqrt{15}}{12+3\sqrt{10}} \cdot \frac{12-3\sqrt{10}}{12-3\sqrt{10}} \\ \\ \\* = \frac{12\sqrt{6}-6\sqrt{15}-144\sqrt{15}+180\sqrt{6}}{144-90} \\ \\ \\* = \frac{192\sqrt{6}-150\sqrt{15}}{54} = \frac{32\sqrt{6}-25\sqrt{15}}{9}$

Report an Error

Example Question #1474 : Pre Calculus

Given that $\tan x = 2$ and $\tan y = 3$ , find $\tan (x+y)$ .

Possible Answers:

Correct answer:

Explanation:

Jump straight to the tangent sum formula:

$\\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}$

From here plug in the given values and simplify.

$\\ \\ = \frac {2+3}{1-2\cdot3} \\ \\= \frac {5}{-5} \\ \\= -1$

Report an Error

Example Question #1475 : Pre Calculus

Which of the following expressions best represents $\tan\left ( 60^{\circ}\right )$ ?

Possible Answers:

$\frac{2\tan(15^\circ)}{1-\tan^2(15^\circ)}$

$\frac{\tan(60^\circ)}{1-\tan^2(60^\circ)}$

$\frac{\tan(30^\circ)}{1-\tan^2(30^\circ)}$

$\frac{2\tan(30^{\circ})}{1-\tan^2(30^{\circ})}$

$\frac{2\tan(60^\circ)}{1-\tan^2(60^\circ)}$

Correct answer:

$\frac{2\tan(30^{\circ})}{1-\tan^2(30^{\circ})}$

Explanation:

Write the identity for $tan(2\theta)$ .

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

Set the value of the angle equal to .

2x=60

x=30

Substitute the value of into the identity.

$\frac{2tan(30)}{1-tan^2(30)}$

Report an Error

Example Question #32 : Trigonometric Identities

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

Possible Answers:

$\sqrt2$

$\frac{\sqrt2}{2}$

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

$\sqrt6+\sqrt2$

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

View Pre-Calculus Tutors

Tajrin
Certified Tutor

CUNY Brooklyn College, Bachelor of Science, Computer Science.

View Pre-Calculus Tutors

Tracy
Certified Tutor

Clarion University of Pennsylvania, Bachelor of Science, Physics.

View Pre-Calculus Tutors

Yaakoub
Certified Tutor

CUNY Queens College, Bachelor of Science, Finance. Vaughn College of Aeronautics and Technology, Bachelor of Science, Electri...

All Precalculus Resources

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

Popular Subjects

Math Tutors in Seattle, GRE Tutors in Los Angeles, French Tutors in San Francisco-Bay Area, MCAT Tutors in New York City, ACT Tutors in Atlanta, ACT Tutors in Dallas Fort Worth, Algebra Tutors in Chicago, Algebra Tutors in Dallas Fort Worth, English Tutors in New York City, Physics Tutors in San Francisco-Bay Area

Popular Courses & Classes

GMAT Courses & Classes in Los Angeles, ACT Courses & Classes in Atlanta, MCAT Courses & Classes in Phoenix, SAT Courses & Classes in San Diego, SSAT Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Boston, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Washington DC, SSAT Courses & Classes in Miami

Popular Test Prep

GMAT Test Prep in Boston, MCAT Test Prep in Atlanta, GRE Test Prep in Houston, ISEE Test Prep in Denver, SAT Test Prep in Seattle, LSAT Test Prep in New York City, ACT Test Prep in Atlanta, SAT Test Prep in Boston, LSAT Test Prep in Houston, ISEE Test Prep in Phoenix

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 : Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1471 : Pre Calculus

Example Question #1472 : Pre Calculus

Example Question #31 : Trigonometric Identities

Example Question #1475 : Pre Calculus

Example Question #1474 : Pre Calculus

Example Question #1475 : Pre Calculus

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: