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

Example Question #1 : Trigonometric Identities

Simplify $\frac{1}{1-\cos^2\theta}$ .

Possible Answers:

$\frac{1}{1-\cos^2\theta}$

$\sec^2\theta$

$\sin\theta\cos\theta$

$\csc^2\theta$

Correct answer:

$\csc^2\theta$

Explanation:

To simplify, recognize that $1-\cos^2\theta$ is a reworking on $\sin^2+\cos^2=1$ , meaning that $1-\cos^2\theta=\sin^2\theta$ .

Plug that into our given equation:

$\frac{1}{1-\cos^2\theta}=\frac{1}{\sin^2\theta}$

Remember that $\csc\theta=\frac{1}{\sin\theta}$ , so $\frac{1}{\sin^2\theta}=\csc^2\theta$ .

Report an Error

Example Question #2 : Trigonometric Identities

Simplify $\frac{1-\cos^2\theta}{\sin\theta\cos\theta}$ .

Possible Answers:

$\cot\theta$

$\cos\theta$

$\sec^2\theta$

$\tan\theta$

$\csc\theta-\cos\theta$

Correct answer:

$\tan\theta$

Explanation:

Recognize that $1-\cos^2\theta$ is a reworking on $\sin^2+\cos^2=1$ , meaning that $1-\cos^2\theta=\sin^2\theta$ .

Plug that in to our given equation:

$\frac{1-\cos^2\theta}{\sin\theta\cos\theta}=\frac{\sin^2\theta}{\sin\theta\cos\theta}$

Notice that one of the $\sin\theta$ 's cancel out.

$\frac{\sin\theta}{\cos\theta}=\tan\theta$ .

Report an Error

Example Question #3 : Pythagorean Identities

$\frac{1}{(cos (33^{\circ}))^{2}} - (tan (33^{\circ}))^{2} = ?$

Possible Answers:

$\frac{5\sqrt{2}}{2}$

$3\sqrt{3}$

-1

Correct answer:

Explanation:

$\frac{1}{(cos (33^{\circ}))^{2}} - (tan (33^{\circ}))^{2} = (sec (33^{\circ}))^{2}- (tan (33^{\circ}))^{2}$

Recall the Pythagorean Identity:

$1 + (tan x)^{2} = (sec x)^{2}$

We can rearrange the terms:

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

This is exactly what our original equation looks like, so the answer is 1.

Report an Error

Example Question #3 : Pythagorean Identities

Simplify the equation using identities:

$\cot^{2}x\cos x \sin x+ \cos x \sin x$

Possible Answers:

$1-\csc x$

$\cos^{2}x$

$\sin x$

$\cot x$

Correct answer:

$\cot x$

Explanation:

There are a couple valid strategies for solving this problem. The simplest is to first factor out $\cos x\sin x$ from both sides. This leaves us with:

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

Next, substitute with the known identity $\cot^{2}x + 1=\csc^{2}x$ to get:

$\cos x\sin x (\csc^{2}x)$

From here, we can eliminate the quadratic by converting:

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

giving us

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

Thus,

$\cot^{2}x\cos x\sin x+\cos x\sin x = \cot x$

Report an Error

Example Question #3 : Trigonometric Identities

Simplify the expression: $(1+\cos a)(1-\cos a)$

Possible Answers:

$1-\cos^{2}a$

$\cos a$

The equation cannot be further simplified.

$\sin a$

$\sin^{2}a$

Correct answer:

$\sin^{2}a$

Explanation:

The expression $(1-\cos a)(1+\cos a)$ represents a difference of squares. In this case, the product is $1-\cos^{2}a$ (remember that 1 is also a perfect square).

One Pythagoran identity for trigonometric functions is:

$1-\cos^{2}a = \sin^{2}a$

Thus, we can say that the most simplified version of the expression is $\sin^{2}a$ .

Report an Error

Example Question #3 : Pythagorean Identities

If theta is in the second quadrant, and $sin(\theta)= \frac{3}{5}$ , what is $cos(\theta)$ ?

Possible Answers:

$-\frac{3}{5}$

$\frac{5}{3}$

$\frac{4}{5}$

$-\frac{4}{5}$

$\frac{3}{5}$

Correct answer:

$-\frac{4}{5}$

Explanation:

Write the Pythagorean Identity.

$sin^2(\theta)+cos^2(\theta)=1$

Substitute the value of $sin(\theta)$ and solve for $cos(\theta)$ .

$(\frac{3}{5})^2+cos^2(\theta)=1$

$cos^2(\theta)=1-(\frac{3}{5})^2$

$cos^2(\theta)=1-(\frac{3}{5})^2= \frac{25}{25}-\frac{9}{25}=\frac{16}{25}$

$cos(\theta)=\pm \sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

Since the cosine is in the second quadrant, the correct answer is:

$cos(\theta)=-\frac{4}{5}$

Report an Error

Example Question #1 : Pythagorean Identities

For which values of is the following equation true?

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

Possible Answers:

$0^{\circ}\leq x \leq 180^{\circ}$

$0^{\circ}\leq x \leq 90^{\circ}$

$\mathbb{R}=(-\infty, \infty )$

$0^{\circ}\leq x <360^{\circ}$

$x = 90^{\circ}, 180^{\circ}, 270^{\circ}$

Correct answer:

$\mathbb{R}=(-\infty, \infty )$

Explanation:

According to the Pythagorean identity

sin^2x+cos^2x=1 ,

the right hand side of this equation can be rewritten as sin^2x . This yields the equation

sinxcosxtanx =sin^2x .

Dividing both sides by sinx yields:

cosxtanx=sinx .

Dividing both sides by cosx yields:

$tanx=\frac{sinx}{cosx}$ .

This is precisely the definition of the tangent function; since the domain of tanx consists of all real numbers, the values of which satisfy the original equation also consist of all real numbers. Hence, the correct answer is

$\mathbb{R}=(-\infty, \infty )$ .

Report an Error

Example Question #1 : Pythagorean Identities

$\sin^2x + \cos^2x + \tan^2x =$

Possible Answers:

$\sec^2x$

$\tan^2x-\cot^2x$

$\csc^2x$

$1-\cos^2x$

Correct answer:

$\sec^2x$

Explanation:

By the Pythagorean identity, the first two terms simplify to 1:

$\sin^2x+\cos^2x+\tan^2x=1+\tan^2x$ .

Dividing the Pythagorean identity by $\cos^2x$ allows us to simplify the right-hand side.

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

$\tan^2x+1=\sec^2x$

Report an Error

Example Question #1 : Trigonometry

What is sin^2x+cos^2x equal to?

Possible Answers:

tan(x)

Correct answer:

Explanation:

Step 1: Recall the trigonometric identity that has sine and cosine in it...

sin^2x+cos^2x=1

The sum is equal to 1.

Report an Error

Example Question #2 : Trigonometric Identities

Given $\sin(\theta)=\frac{1}{4}$ , what is $\cos(\theta)$ ?

Possible Answers:

$\cos(\theta)=\pm15/4$

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$

$\cos(\theta)=15/4$

$\cos(\theta)=15/16$

Correct answer:

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$

Explanation:

Using the Pythagorean Identity

$\sin^2+\cos^2=1$ ,

one can solve for $\cos^2(\theta)$ by plugging in (1/4)^2 for $\sin^2$ .

Solving for $\cos^2(\theta)$ , you get it equal to 15/16 .

Taking the square root of both sides will get the correct answer of

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$ .

Report an Error

View Trigonometry Tutors

Mohammad Talha
Certified Tutor

University of Toronto, Mississauga, Bachelor, Chemistry, and Environmental Science.

View Trigonometry Tutors

Joseph
Certified Tutor

Rutgers University-New Brunswick, Bachelor in Arts, Chemistry. Rutgers Graduate School of Biomedical Sciences, Current Grad S...

View Trigonometry Tutors

Vaishnavi
Certified Tutor

SRM University, Bachelor of Technology, Aerospace Engineering. University of Petroleum and Energy Studies, Masters in Busines...

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

LSAT Courses & Classes in Boston, Spanish Courses & Classes in Washington DC, GMAT Courses & Classes in Phoenix, ACT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Seattle, GRE Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Seattle, Spanish Courses & Classes in Miami, ISEE Courses & Classes in San Diego, GMAT Courses & Classes in Los Angeles

Popular Test Prep

SAT Test Prep in Philadelphia, GMAT Test Prep in Philadelphia, ISEE Test Prep in Boston, GMAT Test Prep in Los Angeles, ACT Test Prep in Washington DC, SAT Test Prep in Atlanta, GMAT Test Prep in Boston, GMAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Philadelphia, SSAT Test Prep in New York City

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 #1 : Trigonometric Identities

Example Question #2 : Trigonometric Identities

Example Question #3 : Pythagorean Identities

Example Question #3 : Pythagorean Identities

Example Question #3 : Trigonometric Identities

Example Question #3 : Pythagorean Identities

Example Question #1 : Pythagorean Identities

Example Question #1 : Pythagorean Identities

Example Question #1 : Trigonometry

Example Question #2 : Trigonometric Identities

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: