Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Pre-Calculus

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 #2 : Trigonometric Functions

What is the restriction of the domain of the function given by:

$y=3sec(3(x-\frac{\pi}{6}))$

For all the answer choices below, is any integer.

Possible Answers:

$x\neq \frac{2\pi}{3}+\pi n$

$x\neq \frac{\pi}{3}+\pi n$

$x\neq \frac{\pi}{6}n$

$x\neq \frac{\pi}{3}n$

Correct answer:

$x\neq \frac{\pi}{3}n$

Explanation:

y=sec(x) has restrictions on its domain such that

$x\neq \frac{\pi}{2}+\pi n$ , where is any integer.

To determine the domain for $y=3sec(3(x-\frac{\pi}{6}))$ ,

we equate the terms within the secant function and set them equal to the original domain restriction.

$3(x-\frac{\pi}{6})\neq \frac{\pi}{2}+\pi n$

Solving for ,

$x-\frac{\pi}{6}\neq\frac{\pi}{6}+\frac{\pi}{3}n$

$x\neq\frac{\pi}{3}+\frac{\pi}{3}n=\frac{\pi}{3}n$

The new domain restriction is:

$x\neq \frac{\pi}{3}n$ where is an integer

Report an Error

Example Question #1 : Trigonometric Functions

Solve the following:

$cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

Rewrite $cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$ in terms of sine and cosine functions.

$\frac{cos(\frac{\pi}{2})}{sin(\frac{\pi}{2})}+\frac{sin(\pi)}{cos(\pi)}-\frac{1}{cos(\frac{\pi}{3})}$

Since these angles are special angles from the unit circle, the values of each term can be determined from the x and y coordinate points at the specified angle.

Solve each term and simplify the expression.

$\frac{0}{1}+\frac{0}{-1}-\frac{1}{\frac{1}{2}}= 0+0-2=-2$

Report an Error

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Q1 new

Find the value of .

Possible Answers:

Correct answer:

Explanation:

Using trigonometric relationships, one can set up the equation

$cos(\theta)=\frac{adjacent}{hypotenuse}$

$cos(30^\circ)=\frac{25}{x}$ .

Solving for ,

$\\x\cdot cos(30^\circ)=25\\ \\x=\frac{25}{cos(30^\circ)}\\ \\x=28.8675... \\ \\x=29$

Thus, the answer is found to be 29.

Report an Error

Example Question #7 : Trigonometric Functions

Q2 new

Find the value of .

Possible Answers:

106

120

Correct answer:

106

Explanation:

Using trigonometric relationships, one can set up the equation

$sin(\theta)=\frac{opposite}{hypotenuse}$ .

Plugging in the values given in the picture we get the equation,

$sin(25) = \frac{45}{x}$ .

Solving for ,

$\\x\cdot sin(25)=45\\ \\x=\frac{45}{sin(25)}\\ \\x=106.479...\\ \\x=106$ .

Thus, the answer is found to be 106.

Report an Error

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Find all of the angles that satistfy the following equation:

$sin(\theta )=\frac{\sqrt{3}}{2}$

Possible Answers:

$\theta= \frac{\pi k}{3}$

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

$\theta= \frac{\pi}{3},\frac{2\pi}{3}$

$\theta= \frac{2\pi}{3}+2\pi k$

$\theta= \frac{\pi}{3}+\pi k$

Correct answer:

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

Explanation:

The values of $\theta$ that fit this equation would be:

$\frac{\pi}{3}$ and $\frac{2\pi}{3}$

because these angles are in QI and QII where sin is positive and where

$sin(\theta)=\frac{\sqrt{3}}{2}$ .

This is why the answer

$\theta= \frac{\pi k}{3}$

is incorrect, because it includes inputs that provide negative values such as:

$sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

Thus the answer would be each $2\pi$ multiple of $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ , which would provide the following equations:

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

Report an Error

Example Question #2 : Trigonometric Functions

Evaluate: $sin\left(\frac{\pi}{2}\right)-sin\left(\frac{\pi}{3}\right)+sin\left(\frac{\pi}{4}\right)$

Possible Answers:

$\frac{2-\sqrt5}{2}$

$\frac{2-\sqrt3+\sqrt2}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{2-\sqrt3+\sqrt2}{2}$

Explanation:

To evaluate $sin(\frac{\pi}{2})-sin(\frac{\pi}{3})+sin(\frac{\pi}{4})$ , break up each term into 3 parts and evaluate each term individually.

$sin(\frac{\pi}{2})=1$

$sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$

$sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

Simplify by combining the three terms.

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

Report an Error

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

What is the value of $tan(\pi)+sin(\pi)$ ?

Possible Answers:

$\frac{\sqrt2}{2}$

$\infty$

Correct answer:

Explanation:

Convert $tan(\pi)+sin(\pi)$ in terms of sine and cosine.

$tan(\pi)+sin(\pi)= \frac{sin(\pi)}{cos(\pi)}+sin(\pi)$

Since theta is $\pi$ radians, the value of $sin(\pi)$ is the y-value of the point on the unit circle at $\pi$ radians, and the value of $cos(\pi)$ corresponds to the x-value at that angle.

The point on the unit circle at $\pi$ radians is (-1,0) .

Therefore, $sin(\pi)=0$ and $cos(\pi)=-1$ . Substitute these values and solve.

$\frac{sin(\pi)}{cos(\pi)}+sin(\pi)=\frac{0}{-1}+0=0$

Report an Error

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Solve: $\frac{1}{sin(\frac{\pi}{4})}$

Possible Answers:

$\frac{1}{2}$

$2\sqrt2$

$\sqrt2$

$\frac{\sqrt2}{2}$

Correct answer:

$\sqrt2$

Explanation:

First, solve the value of $sin\left(\frac{\pi}{4}\right)$ .

On the unit circle, the coordinate at $\frac{\pi}{4}$ radians is $\left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)$ . The sine value is the y-value, which is $\frac{\sqrt2}{2}$ . Substitute this value back into the original problem.

$\frac{1}{sin(\frac{\pi}{4})}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}$

Rationalize the denominator.

$\frac{2}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}= \sqrt2$

Report an Error

Example Question #5 : Find The Value Of Any Of The Six Trigonometric Functions

Find the exact answer for: cos(60)-sin(30)+cos(30)

Possible Answers:

$\sqrt3$

$-\frac{\sqrt3}{2}$

$\frac{\sqrt3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

To evaluate cos(60)-sin(30)+cos(30) , solve each term individually.

cos(60) refers to the x-value of the coordinate at 60 degrees from the origin. The x-value of this special angle is $\frac{1}{2}$ .

sin(30) refers to the y-value of the coordinate at 30 degrees. The y-value of this special angle is $\frac{1}{2}$ .

cos(30) refers to the x-value of the coordinate at 30 degrees. The x-value is $\frac{\sqrt3}{2}$ .

Combine the terms to solve cos(60)-sin(30)+cos(30) .

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

Report an Error

Example Question #231 : Pre Calculus

Find the value of

$\sin(300)$ .

Possible Answers:

$-\sqrt2$

$-\frac{\sqrt2}{2}$

$-\frac{\sqrt3}{2}$

$-\frac{1}{2}$

$-\sqrt3}$

Correct answer:

$-\frac{\sqrt3}{2}$

Explanation:

The value of sin(300) refers to the y-value of the coordinate that is located in the fourth quadrant.

This angle 300 is also from the origin.

Therefore, we are evaluating sin(-60) .

$sin(-60)=-sin(60)=-\frac{\sqrt3}{2}$

Report an Error

View Pre-Calculus Tutors

Bobby
Certified Tutor

University of Houston - Downtown, Bachelor of Science, Applied Mathematics.

View Pre-Calculus Tutors

Sheela
Certified Tutor

University of Pittsburgh, Master of Science, Information Science.

View Pre-Calculus Tutors

Zara
Certified Tutor

University of Manitoba, Bachelor in Arts, Mathematics.

All Precalculus Resources

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

Popular Subjects

ISEE Tutors in Seattle, Computer Science Tutors in New York City, Chemistry Tutors in Phoenix, English Tutors in Denver, Reading Tutors in Philadelphia, Computer Science Tutors in Washington DC, Calculus Tutors in Houston, Math Tutors in Phoenix, Chemistry Tutors in Denver, Statistics Tutors in New York City

Popular Courses & Classes

Spanish Courses & Classes in Chicago, SAT Courses & Classes in Denver, GRE Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Los Angeles, Spanish Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in Atlanta, SSAT Courses & Classes in Miami, MCAT Courses & Classes in Washington DC, SSAT Courses & Classes in Denver

Popular Test Prep

GMAT Test Prep in Phoenix, SSAT Test Prep in Phoenix, ISEE Test Prep in Los Angeles, ISEE Test Prep in Seattle, ISEE Test Prep in Dallas Fort Worth, ISEE Test Prep in San Diego, SAT Test Prep in Miami, SSAT Test Prep in Denver, MCAT Test Prep in Atlanta, GMAT 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 : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #7 : Trigonometric Functions

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #2 : Trigonometric Functions

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #5 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #231 : Pre Calculus

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: