Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : Trigonometric Operations

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Trigonometry

If the polar coordinates of a point are $\left (2,\frac{-\pi}{4} \right )$ , then what are its rectangular coordinates?

Possible Answers:

$(2,-\sqrt2)$

$(-\sqrt2,-\sqrt2)$

$(\sqrt2,\sqrt2)$

$(\sqrt2,-\sqrt2)$

(2,2)

Correct answer:

$(\sqrt2,-\sqrt2)$

Explanation:

The polar coordinates of a point are given as $(r,\theta )$ , where r represents the distance from the point to the origin, and $\theta$ represents the angle of rotation. (A negative angle of rotation denotes a clockwise rotation, while a positive angle denotes a counterclockwise rotation.)

The following formulas are used for conversion from polar coordinates to rectangular (x, y) coordinates.

$x=r\cos\theta$

$y=r\sin\theta$

In this problem, the polar coordinates of the point are $\left (2,\frac{-\pi}{4} \right )$ , which means that r = 2 and $\theta=-\frac{\pi}{4}$ . We can apply the conversion formulas to find the values of x and y.

$x=r\cos\theta=2\cos\left (\frac{-\pi}{4} \right )=2\cdot \frac{\sqrt2}{2}= \sqrt{2}$

$y=r\sin\theta=2\sin\left (\frac{-\pi}{4} \right )=2\cdot \frac{-\sqrt2}{2}= -\sqrt{2}$

The rectangular coordinates are $(\sqrt2,-\sqrt2)$ .

The answer is $(\sqrt2,-\sqrt2)$ .

Report an Error

Example Question #1 : Trigonometry

Triangle

What is the $\sin C$ ?

Possible Answers:

$\frac{BC}{AC}$

$\frac{BC}{AB}$

$\frac{AC}{AB}$

$\frac{AB}{AC}$

$\frac{AB}{BC}$

Correct answer:

$\frac{AB}{AC}$

Explanation:

$\sin X = \frac{OPPOSITE}{HYPOTENUSE}$

$\sin C = \frac{AB}{AC}$

Report an Error

Example Question #1 : Trigonometry

Triangle

In the right triangle above, which of the following expressions gives the length of y?

Possible Answers:

$x\cos\theta$

$x\sin\theta$

$\cos(x\times \theta)$

$x\tan\theta$

$\frac{\sin(x)}{\theta}$

Correct answer:

$x\cos\theta$

Explanation:

$\cos\theta$ is defined as the ratio of the adjacent side to the hypotenuse, or in this case $\frac{y}{x}$ . Solving for y gives the correct expression.

Report an Error

Example Question #2 : Understanding Sine, Cosine, And Tangent

What is the cosine of $30^\circ$ ?

Possible Answers:

$\sqrt{3}$

$\frac{1}{3}$

$\frac{1}{2}$

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

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

Correct answer:

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

Explanation:

The pattern for the side of a 30:60:90 triangle is $x:x\sqrt{3}:2x$ .

Since $\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$ , we can plug in our given values.

$\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$

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

Notice that the 's cancel out.

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

Report an Error

Example Question #3 : Understanding Sine, Cosine, And Tangent

If sin(x)=0.5 , what is sin(8x) if is between and $2\pi$ ?

Possible Answers:

0.5

-1.5

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

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

-0.5

Correct answer:

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

Explanation:

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .

Report an Error

Example Question #2 : Trigonometry

An angle has a cosine of $\frac{4}{5}$ . What will its cosecant be?

Possible Answers:

$\frac{4}{3}$

$\frac{5}{4}$

$\frac{3}{4}$

$\frac{5}{3}$

$\frac{3}{5}$

Correct answer:

$\frac{5}{3}$

Explanation:

The problem tells us that the cosine of the angle will be $\frac{4}{5}$ . Cosine is the adjacent over the hypotenuse. From here we can use the Pythaogrean theorem:

$\text{Opposite}^2+\text{Adjacent}^2=\text{Hypotenuse}^2$

O^2+4^2=5^2

O^2+16=25

O^2=9

$\sqrt{O^2}=\sqrt{9}$

O=3

Now we know our opposite, adjacent, and hypotenuse.

The cosecant is $\csc=\frac{\text{Hypotenuse}}{\text{Opposite}}$ .

From here we can plug in our given values.

$\csc=\frac{5}{3}$

Report an Error

Example Question #1 : Trigonometry

Which of these is equal to $\frac{\text{hypotenuse}}{\text{adjacent}}$ for angle ?

Possible Answers:

$\cos{x}$

$\csc{x}$

$\sec{x}$

$\cot{x}$

$\sin x$

Correct answer:

$\sec{x}$

Explanation:

$\sec{x}=\frac{\text{hypotenuse}}{\text{adjacent}}$ , as it is the inverse of the $\cos{x}$ function. This is therefore the answer.

Report an Error

Example Question #1 : Trigonometry

Trig_id

What is $\theta$ if o=10 and a=8 ?

Possible Answers:

$53.14^{\circ}$

$55.14^{\circ}$

$51.34^{\circ}$

$0.90^{\circ}$

$89.60^{\circ}$

Correct answer:

$51.34^{\circ}$

Explanation:

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

Report an Error

Example Question #1 : Trigonometry

$\sin(x^\circ)=\frac{1}{2}$

What is ?

Possible Answers:

$60^\circ$

$30^\circ$

$26.57^\circ$

$0.009^\circ$

$1^\circ$

Correct answer:

$30^\circ$

Explanation:

To get rid of $\sin(x^\circ)$ , we take the $\arcsin$ or $\sin^{-1}$ of both sides.

$\sin(x^\circ)=\frac{1}{2}$

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

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

$x^\circ=30^\circ$

Report an Error

Example Question #4 : Trigonometry

Trig_id

$o=7.6, h=15.7, \Theta = ?$

Possible Answers:

$0.50^{\circ}$

$85.92^{\circ}$

$28.95^{\circ}$

$59.28^{\circ}$

$50.53^{\circ}$

Correct answer:

$28.95^{\circ}$

Explanation:

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and hypotenuse sides. We can then, by definition, find the $\sin$ of $\theta$ and its measure in degrees by utilizing the $\arcsin$ function.

$\sin=\frac{opposite}{hypotenuse}$

$\sin=\frac{7.6}{15.7}$

$\sin=0.4841$

Now to find the measure of the angle using the $\arcsin$ function.

$\theta=\arcsin0.4841$

$\rightarrow 28.95^{\circ}$

If you calculated the angle's measure to be $\dpi{100} 0.50^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

Report an Error

View Tutors

Travis
Certified Tutor

Delaware State University, Bachelor of Science, Mathematics.

View Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

View Tutors

Ronald
Certified Tutor

Towson University, Master of Science, Counseling Psychology.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Phoenix, Algebra Tutors in Washington DC, English Tutors in Boston, Algebra Tutors in Houston, Computer Science Tutors in Dallas Fort Worth, English Tutors in Washington DC, English Tutors in Phoenix, ISEE Tutors in Washington DC, Reading Tutors in Houston, GRE Tutors in Seattle

Popular Courses & Classes

SAT Courses & Classes in San Diego, SSAT Courses & Classes in Atlanta, LSAT Courses & Classes in San Diego, GMAT Courses & Classes in New York City, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Denver, ACT Courses & Classes in Phoenix, ACT Courses & Classes in Washington DC, ACT Courses & Classes in Los Angeles, GMAT Courses & Classes in Chicago

Popular Test Prep

GMAT Test Prep in Chicago, ISEE Test Prep in San Francisco-Bay Area, MCAT Test Prep in Chicago, ACT Test Prep in Los Angeles, ACT Test Prep in Phoenix, GRE Test Prep in Seattle, MCAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in San Diego, ACT Test Prep in Washington DC, GRE 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

High School Math : Trigonometric Operations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #2 : Understanding Sine, Cosine, And Tangent

Example Question #3 : Understanding Sine, Cosine, And Tangent

Example Question #2 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #4 : Trigonometry

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: