We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Trigonometry : Trigonometry

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

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 76 77 Next →

Example Question #81 : Trigonometry

If $x=\frac{\pi}{6}$ , find the value of $\frac{1+cos\ x}{sin^3x}$ .

Possible Answers:

$-8+4\sqrt{3}$

$8-4\sqrt{3}$

$8+4\sqrt{2}$

$4+4\sqrt{3}$

$8+4\sqrt{3}$

Correct answer:

$8+4\sqrt{3}$

Explanation:

$x=\frac{\pi}{6}\Rightarrow \frac{{1+cos\ x}}{}{sin^3x}=\frac{{1+cos\ (\pi/6)}}{sin^3(\pi/6)}=\frac{1+\frac{\sqrt{3}}{2}}{(\frac{1}{2})^3}=4(2+\sqrt{3})=8+4\sqrt{3}$

Report an Error

Example Question #82 : Trigonometry

Find the value of the following expression:

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}$

Possible Answers:

Correct answer:

Explanation:

We know that $sec\ x= \frac{1}{cos\ x}$ .

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}=sin^2 (\frac{\pi}{15})+\frac{1}{\frac{1}{cos^2(\frac{\pi}{15})}}=sin^2 (\frac{\pi}{15})+cos^2 (\frac{\pi}{15})=1$

Report an Error

Example Question #81 : Trigonometry

If $sec\ x=\frac{\sqrt{5}}{2}$ , give $tan\ x$ :

Possible Answers:

$tan\ x=\pm 2$

$tan\ x=\pm 3$

$tan\ x=\pm 4$

$tan\ x=\pm 1$

$tan\ x=\pm \frac{1}{2}$

Correct answer:

$tan\ x=\pm \frac{1}{2}$

Explanation:

We need to use the identity sec^2x=1+tan^2x :

$sec^2x=1+tan^2x\Rightarrow (\frac{\sqrt{5}}2)^2=1+tan^2x$

$\Rightarrow tan^2x=\frac{5}{4}-1\Rightarrow tan^2x=\frac{1}{4}$

$\Rightarrow tan\ x=\pm \frac{1}{2}$

Report an Error

Example Question #83 : Trigonometry

Give the value of $sec\ 26^{\circ} - csc\ 64^{\circ}$ .

Possible Answers:

Correct answer:

Explanation:

We need to use the identity $csc(90^{\circ}-x)=sec\ x$ :

$sec\ 26^{\circ} - csc\ 64^{\circ}=sec\ 26^{\circ} - csc (90-26)^{\circ}=sec\ 26^{\circ} -sec\ 26^{\circ}=0$

Report an Error

Example Question #84 : Trigonometry

Find $\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}$ .

Possible Answers:

$sin\ 25^{\circ}$

$-sin\ 25^{\circ}$

Correct answer:

Explanation:

We need to use the identity $cos(90^{\circ}-x)=sin\ x$ :

$\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}=\frac{cos (90-25)^{\circ}}{sin\ 25^{\circ}}=\frac{sin\ 25^{\circ}}{sin\ 25^{\circ}}=1$

Report an Error

Example Question #1 : Trigonometric Functions And Graphs

If $cos\ x=A$ , give $cos\ 2x$ in terms of .

Possible Answers:

A-1

A+1

2A^2+1

2A^2-1

A^2-1

Correct answer:

2A^2-1

Explanation:

We need to use the identity cos2x=2cos^2x-1 .

cos2x=2cos^2x-1=2A^2-1

Report an Error

Example Question #85 : Trigonometry

If $cos\ x=\frac{\sqrt{2}}{2}$ , find $cos\ 2x$ .

Possible Answers:

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

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

Correct answer:

Explanation:

We need to use the identity cos2x=2cos^2x-1 .

$cos2x=2cos^2x-1=2(\frac{\sqrt{2}}{2})^2-1=2(\frac{1}{2})-1=1-1=0$

Report an Error

Example Question #11 : Trigonometric Functions

If $sin\ x=A$ , express $sin\ 2x$ in terms of . ( $0<x<30^{\circ}$ )

Possible Answers:

$2\sqrt{1-A^2}$

$2A\sqrt{1+A^2}$

$A\sqrt{1+A^2}$

$A\sqrt{1-A^2}$

$2A\sqrt{1-A^2}$

Correct answer:

$2A\sqrt{1-A^2}$

Explanation:

We need to use the double angle formula:

$sin\ 2x=2(sin\ x)(cos\ x)$

$sin\ x$ is known, but we need to find $cos \ x$ :

$sin^2x+cos^2x=1\Rightarrow cos\ x=\pm \sqrt{1-sin^2x}$

In this problem is a first quadrant angle ( $0<x<30^{\circ}$ ), so we can only use the positive value for $cos\ x$ .

$cos\ x=\sqrt{1-sin^2x}=\sqrt{1-A^2}$

$sin\ 2x=2(sin\ x)(cos\ x)=2A\sqrt{1-A^2}$

Report an Error

Example Question #11 : Trigonometric Functions And Graphs

If $tan\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Possible Answers:

0.75

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

0.5

0.25

Correct answer:

0.5

Explanation:

We need to use a Pythagorean identity:

$sec^2x=1+tan^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow sec\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value of $sec\ x$ . That means $sec\ x=2$ .

$cos\ x=\frac{1}{sec\ x}=\frac{1}{2}=0.5$

Report an Error

Example Question #81 : Trigonometry

If $cot\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Possible Answers:

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

$\frac{1}{2}$

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

Correct answer:

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

Explanation:

We need to use a Pythagorean identity:

$csc^2x=1+cot^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow csc\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value for $csc\ x$ . That means $csc\ x=2$ .

$sin\ x=\frac{1}{csc\ x}=\frac{1}{2}$

Based on another Pythagorean identity we have:

$cos^2x+sin^2x=1\Rightarrow cos^2x=1-sin^2x=1-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$

$\Rightarrow cos\ x=\pm \sqrt{(\frac{3}{4})}=\pm \frac{\sqrt{3}}{2}$

Since $0^{\circ}<x<90^{\circ}$ , we can again only use the positive value for $cos\ x$ .

$cos\ x=\frac{\sqrt{3}}{2}$

Report an Error

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 76 77 Next →

View Trigonometry Tutors

Sekou
Certified Tutor

University of New Haven, Bachelor in Arts, Mathematics.

View Trigonometry Tutors

Kaitlyn
Certified Tutor

Fairfield University, Bachelor of Science, Biology, General. NUI Galway Ireland, Master of Science, Neuroscience.

View Trigonometry Tutors

Liban
Certified Tutor

Emory University, Bachelor of Science, Mathematics/Economics.

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

ACT Courses & Classes in Phoenix, SSAT Courses & Classes in Washington DC, GRE Courses & Classes in New York City, ISEE Courses & Classes in Houston, Spanish Courses & Classes in Seattle, GRE Courses & Classes in Atlanta, LSAT Courses & Classes in Chicago, MCAT Courses & Classes in Washington DC, MCAT Courses & Classes in Miami, Spanish Courses & Classes in Washington DC

Popular Test Prep

ACT Test Prep in Boston, MCAT Test Prep in Miami, LSAT Test Prep in Chicago, GRE Test Prep in Seattle, GMAT Test Prep in Los Angeles, SSAT Test Prep in Los Angeles, ISEE Test Prep in Houston, GRE Test Prep in Denver, SAT Test Prep in Boston, GRE Test Prep in Dallas Fort Worth

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 : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #81 : Trigonometry

Example Question #82 : Trigonometry

Example Question #81 : Trigonometry

Example Question #83 : Trigonometry

Example Question #84 : Trigonometry

Example Question #1 : Trigonometric Functions And Graphs

Example Question #85 : Trigonometry

Example Question #11 : Trigonometric Functions

Example Question #11 : Trigonometric Functions And Graphs

Example Question #81 : Trigonometry

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: