Call Now to Set Up Tutoring:

(888) 888-0446

Trigonometry : Trigonometric Functions and Graphs

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 #1 : Simplifying Trigonometric Functions

If $cos\ x=\frac{1}{3}$ and $sin\ x\neq 0$ , give the value of $\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$ .

Possible Answers:

Correct answer:

Explanation:

Based on the double angle formula we have, $sin\ 2x=2(sin\ x)(cos\ x)$ .

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

$=\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}=\frac{1+2cos\ x}{1-2cos\ x}$

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

Report an Error

Example Question #1 : Simplifying Trigonometric Functions

If $x=\frac{\pi}{3}$ , give the value of $\frac{sin\ x}{sin\ x+cos\ 2x}$ .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$x=\frac{\pi}{3}\Rightarrow sin\ x=sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Now we can write:

$x=\frac{\pi}{3}\Rightarrow 2x=2(\frac{\pi}{3})=\frac{2\pi}{3}\Rightarrow cos\ 2x=cos(\frac{2\pi}{3})$

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

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

$=-\frac{1}{2}$

Now we can substitute the values:

$\frac{sin\ x}{sin\ x+cos\ 2x}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}=\frac{3+\sqrt{3}}{2}$

Report an Error

Example Question #101 : Trigonometry

If $cot\ 34^{\circ}=1.5$ , give the value of $\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}$ .

Possible Answers:

1.5

0.5

2.5

Correct answer:

2.5

Explanation:

$326^{\circ}=360^{\circ}-34^{\circ}\Rightarrow sin\ 326^{\circ}=sin(360^{\circ}-34^{\circ})\Rightarrow sin\ 326^{\circ}=-sin\ 34^{\circ}$

$56^{\circ}=90^{\circ}-34^{\circ}\Rightarrow sin\ 56^{\circ}=sin(90^{\circ}-34^{\circ})\Rightarrow sin\ 56^{\circ}=cos\ 34^{\circ}$

$304^{\circ}=270^{\circ}+34^{\circ}\Rightarrow cos\ 304^{\circ}=cos(270^{\circ}+34^{\circ})\Rightarrow cos\ 304^{\circ}=sin\ 34^{\circ}$

Now we can simplify the expression as follows:

$\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}=\frac{-2sin\ 34^{\circ}+3cos\ 34^{\circ}}{sin\ 34^{\circ}}$

$=-2+3cot\ 34^{\circ}=-2+3\times 1.5=-2+4.5=2.5$

Report an Error

Example Question #11 : Simplifying Trigonometric Functions

If $sin\ x\times cos\ x=\frac{1}{2}$ , what is the value of $(sin\ x+cos\ x)^2$ ?

Possible Answers:

1.5

2.5

Correct answer:

Explanation:

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

$=1+2(sin\ x)(cos\ x)=1+2(\frac{1}{2})=1+1=2$

Report an Error

Example Question #33 : Trigonometric Functions And Graphs

If $sin^2x-cos^2x=\frac{1}{2}$ , give the value of cos^4x-sin^4x .

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x)

We know that cos^2x+sin^2x=1 .

cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x)

$=(cos^2x-sin^2x)\times 1=-(sin^2x-cos^2x)=-\frac{1}{2}$

Report an Error

Example Question #12 : Simplifying Trigonometric Functions

Simplify:

A=(cos^4x-sin^4x)(1+tan^2x)+tan^2x

Possible Answers:

1.5

2.5

0.5

Correct answer:

Explanation:

A=(cos^4x-sin^4x)(1+tan^2x)+tan^2x

= (cos^2x-sin^2x)(cos^2x+sin^2x)(1+tan^2x)+tan^2x

We know that cos^2x+sin^2x=1 .

Then we can write:

A= (cos^2x-sin^2x)(1+tan^2x)+tan^2x

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

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

$\Rightarrow A=\frac{(cos^2x-sin^2x)}{cos^2x}+tan^2x$

$=\frac{cos^2x}{cos^2x}-\frac{sin^2x}{cos^2x}+tan^2x$

=1-tan^2x+tan^2x

Report an Error

Example Question #13 : Simplifying Trigonometric Functions

Which of the following is equal to $\sin x\tan x$ ?

Possible Answers:

$\frac{\sin^2x}{\cos x}$

$\sec x \csc x$

$\frac{\cos^2x}{\sin x}$

$\cos x \cot x$

$\sin x \csc x$

Correct answer:

$\frac{\sin^2x}{\cos x}$

Explanation:

Break $\tan x$ apart: $\tan x =\frac{\sin x }{\cos x}$ .

This means that $\sin x\tan x=\sin x *\frac{\sin x}{\cos x}$ or $\frac{\sin^2 x}{\cos x}$

Report an Error

Example Question #12 : Simplifying Trigonometric Functions

Simplify: $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$

Possible Answers:

$cos(\theta)+cot(\theta)$

$sec(\theta)-tan(\theta)$

$cos(\theta)-sin(\theta)cos(\theta)$

$cos(\theta)cot(\theta)-cot(\theta)$

$cos(\theta)-cot(\theta)$

Correct answer:

$cos(\theta)-cot(\theta)$

Explanation:

Rewrite $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$ in terms of sines and cosines.

$\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)} = \frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}$

Simplify the complex fractions.

$\frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}=sin(\theta)\cdot \frac{cos(\theta)}{sin(\theta)}-\frac{1}{sin(\theta)}\cdot \frac{cos(\theta)}{1}$

$=cos(\theta)-\frac{cos(\theta)}{sin(\theta)}= cos(\theta)-cot(\theta)$

Report an Error

Example Question #14 : Simplifying Trigonometric Functions

Simplify the following expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

Possible Answers:

tan^2x

cot^2x

$cos \ 2x$

cos^2x

Correct answer:

cot^2x

Explanation:

We will first invoke the appropriate ratio for cotangent, and then use pythagorean identities to simplify the expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

$= \frac{1}{cos^2x} * \frac{cos^2}{sin^2x} - (sec^2 x - tan^2x){}$

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

$= csc^2x - 1{}$

since

$sec^2x - tan^2x = 1; \ csc\ x = \frac{1}{sin \ x}{}$

Report an Error

Example Question #15 : Simplifying Trigonometric Functions

Simplify the trigonometric expression.

$\frac{\tan(\theta)\csc(\theta)}{\sec(\theta)}$

Possible Answers:

$\cos(\theta)$

$\sec\theta$

$\tan(\theta)$

Correct answer:

Explanation:

Using basic trigonometric identities, we can simplify the problem to

$\frac{\frac{\sin(\theta)}{\cos(\theta)}*\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}$ .

We can cancel the sine in the numerator and the one over cosine cancels on top and bottom, leaving us with 1.

$\frac{\frac{\sin(\theta)}{\cos(\theta)}*\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{\frac{1}{\cos(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{1}{\cos(\theta)}\cdot \frac{\cos(\theta)}{1}=1$

Report an Error

View Trigonometry Tutors

Ivory
Certified Tutor

Hopkinsville Community College, Associate, Associates in Science. Western Kentucky University, Bachelor in Arts, Psychology.

View Trigonometry Tutors

Ana Maria
Certified Tutor

Florida International University, Bachelor of Science, Mathematics Teacher Education.

View Trigonometry Tutors

Kevin
Certified Tutor

St. Johns University, Bachelors, Math/Coaching.

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

SSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Miami, LSAT Courses & Classes in Boston, MCAT Courses & Classes in Miami, GRE Courses & Classes in Chicago, SSAT Courses & Classes in Atlanta, ISEE Courses & Classes in Seattle, ISEE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Miami, GMAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

GRE Test Prep in Philadelphia, GRE Test Prep in Atlanta, SAT Test Prep in Houston, ACT Test Prep in Chicago, ISEE Test Prep in Miami, ISEE Test Prep in Washington DC, ISEE Test Prep in Seattle, MCAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in San Diego, GMAT 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

Trigonometry : Trigonometric Functions and Graphs

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Simplifying Trigonometric Functions

Example Question #1 : Simplifying Trigonometric Functions

Example Question #101 : Trigonometry

Example Question #11 : Simplifying Trigonometric Functions

Example Question #33 : Trigonometric Functions And Graphs

Example Question #12 : Simplifying Trigonometric Functions

Example Question #13 : Simplifying Trigonometric Functions

Example Question #12 : Simplifying Trigonometric Functions

Example Question #14 : Simplifying Trigonometric Functions

Example Question #15 : Simplifying Trigonometric Functions

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: