Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Trigonometric Functions

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 #1502 : Pre Calculus

Compute

$\small 2\cos^2 \frac{\pi}{8}$

Possible Answers:

$\small \frac{\sqrt{2}}{2}+1$

$\small \pi-2$

$\small 0$

$\small \frac{\sqrt{3}}{2}-1$

Correct answer:

$\small \frac{\sqrt{2}}{2}+1$

Explanation:

A useful trigonometric identity to remember for this problem is

$\small 2\cos^2 \theta-1=\cos2\theta$

or equivalently,

$\small \small 2\cos^2 \theta=1+\cos2\theta$

If we substitute $\small \theta$ for $\small \frac{\pi}{8}$ , we get

$\small \small \small 2\cos^2 \frac{\pi}{8}=1+\cos2\frac{\pi}{8}=1+\cos\frac{\pi}{4}=1+\frac{\sqrt{2}}{2}$

Report an Error

Example Question #1502 : Pre Calculus

Compute

$\small \cos\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{8}\right)$

Possible Answers:

$\small \small \frac{\sqrt{6}}{9}$

$\small \frac{\sqrt{3}}{4}$

$\small \small \frac{\sqrt{2}}{4}$

$\small \frac{1+\pi}{\sqrt{5}}$

Correct answer:

$\small \small \frac{\sqrt{2}}{4}$

Explanation:

A useful trigonometric identity to remember is

$\small 2\cos x \sin x = \sin2x$

If we plug in $\small \pi/8$ into this equation, we get

$\small \small 2\cos \left(\frac{\pi}{8}\right)\sin \left(\frac{\pi}{8}\right) = \sin2\left(\frac{\pi}{8}\right)=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$

We can divide the equation by 2 to get

$\small \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{4}$

Report an Error

Example Question #1503 : Pre Calculus

Using the half-angle identities, which of the following answers best resembles sin(30) ?

Possible Answers:

$\pm\sqrt{\frac{1-cos(15)}{2}}$

$\pm\sqrt{\frac{1+cos(15)}{2}}$

$\pm\sqrt{\frac{1-cos(60)}{2}}$

$\pm\sqrt{\frac{1+sin(60)}{1-sin(60)}}$

$\pm\sqrt{\frac{1+sin(30)}{2}}$

Correct answer:

$\pm\sqrt{\frac{1-cos(60)}{2}}$

Explanation:

Write the half angle identity for sine.

$sin\left(\frac{A}{2}\right)= \pm\sqrt{\frac{1-cosA}{2}}$

Since we are given sin(30) , the angle is equal to $\frac{A}{2}$ . Set these two angles equal to each other and solve for .

$\frac{A}{2}=30$

A=60

Substitute this value into the formula.

$sin\left(\frac{60}{2}\right)= \pm\sqrt{\frac{1-cos(60)}{2}}$

Report an Error

Example Question #23 : Fundamental Trigonometric Identities

Let and two reals. Given that:

$$sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$$

$$sin(a-b)=sin(a)cos(b)-cos(a)sin(b)$$

What is the value of:

$sin^2(a+b)-sin^2(a-b)$ ?

Possible Answers:

sin(2a)sin(2b)

2sin(2a)sin(2b)

cos(2a)sin(2b)

sin(2a)cos(2b)

sin(a)sin(b)

Correct answer:

sin(2a)sin(2b)

Explanation:

We have:

sin^2(a+b)=(sin(a)cos(b)+cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)+2sin(a)cos(b)cos(a)sin(b)(1)$$

and :

sin^2(a-b)=(sin(a)cos(b)-cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)-2sin(a)cos(b)cos(a)sin(b)(2)$$

(1)-(2) gives:

4sin(a)cos(b)cos(a)sin(b)=[2sin(a)cos(a)] [2sin(b)cos(b)]$$

Knowing from the above formula that:( take a=b in the formula above)

sin(2a)=2sin(a)cos(a) and sin(2b)=2sin(b)cos(b)$

This gives:

sin^2(a+b)-sin^2(a-b)=sin(2a)sin(2b)$

Report an Error

Example Question #144 : Trigonometric Functions

Let , , and be real numbers. Given that:

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

sin(a-b)=sin(a)cos(b)-cos(a)sin(b)

What is the value of sin(3a) in function of sin(a) ?

Possible Answers:

3sin(a)-4sin^3(a)

3sin(a)-4sin^4(a)

4sin(a)-3sin^3(a)

6sin(a)-4sin^3(a)

3sin(2a)-4sin^3(a)

Correct answer:

3sin(a)-4sin^3(a)

Explanation:

We note first, using trigonometric identities that:

sin(3a)=sin(a+2a)=sin(a)cos(2a)+cos(a)sin(2a)

cos(2a)=1-2sin^2(a)

sin(2a)=2sin(a)cos(a)

This gives:

sin(3a)=sin(a)(1-2sin^2(a))+2sin(a)cos^2(a)

Since, cos^2(a)=1-sin^2(a)

We have :

sin(3a)=sin(a)-2sin^3(a)+2sin(a)-2sin^3(a)=3sin(a)-4sin^3(a)

Report an Error

Example Question #146 : Trigonometric Functions

Using the fact that,

sin(x+180)=-sin(x) .

What is the result of the following sum:

$sin(1)+sin(2)+sin(3)\cdots +sin(359)$

Possible Answers:

sin(1)

cos(1)

Correct answer:

Explanation:

We can write the above sum as :

$sin(1)+sin(181)+sin(2)+sin(182)\cdots$

+ sin(179)+sin(379) + sin(180)

From the given fact, we have :

sin(1)+sin(181)=sin(1)+sin(1+180)=sin(1)-sin(1)=0

sin(2)+sin(182)=sin(2)+sin(2+180)=sin(2)-sin(2)=0

$\vdots$

sin(179)+sin(379)=sin(179)+sin(179+180)=sin(179)-sin(179)=0

and we have : sin(180)=0 .

This gives :

$sin(1)+sin(2)+sin(3)\cdots +sin(359)=0$

Report an Error

Example Question #31 : Fundamental Trigonometric Identities

Compute cos(4x) in function of sin(x) .

Possible Answers:

1-8sin^2(x)+6sin^4(x)

1-8sin^2(x)+8sin^4(x)

1+8sin^2(x)+8sin^4(x)

1-8sin^2(x)-8sin^4(x)

1-6sin^2(x)+8sin^4(x)

Correct answer:

1-8sin^2(x)+8sin^4(x)

Explanation:

Using trigonometric identities we have :

cos(4x)=1-2sin^2(2x) and we know that:

sin(2x)=2sin(x)cox(x)

This gives us :

cos(4x)=1-2(2sin(x)cox(x))^2=1-8sin^2(x)cos^2(x)

= 1-8sin^2(x)(1-sin^2(x)= 1-8sin^2(x)+8sin^4(x)

Hence:

cos(4x)=1-8sin^2(x)+8sin^4(x)

Report an Error

Example Question #31 : Fundamental Trigonometric Identities

Given that :

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

Let,

$tan\left(\frac{x}{8}\right)=s$

What is $tan\left(\frac{x}{4}\right)$ in function of ?

Possible Answers:

$\frac{4s}{1-4s^2}$

$\frac{s}{1-4s^2}$

$\frac{2s}{1-s^2}$

$\frac{4s}{1-s^2}$

$\frac{2s}{1-2s^2}$

Correct answer:

$\frac{2s}{1-s^2}$

Explanation:

We will use the given formula :

We have in this case:

$tan\left(\frac{x}{4}\right)=\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}$

Since we know that :

$tan\left(\frac{x}{8}\right)=s$

This gives :

$\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}=\frac{2s}{1-s^2}$

Report an Error

Example Question #31 : Fundamental Trigonometric Identities

Using the fact that cos(x+180)=-cos(x) , what is the result of the following sum:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)$

Possible Answers:

Correct answer:

Explanation:

We can write the above sum as :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

From the given fact, we have :

cos(1)+cos(181)=cos(1)+cos(1+180)=cos(1)-cos(1)=0

cos(2)+cos(182)=cos(2)+cos(2+180)=cos(2)-cos(2)=0

$\vdots$

cos(179)+cos(379)=cos(179)+cos(179+180)=cos(179)-cos(179)=0

This gives us :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

=0+cos(180)=-1

Therefore we have:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)=-1$

Report an Error

Example Question #147 : Trigonometric Functions

Let $\alpha, \beta$ be real numbers. If $tan(2x)=\alpha$ and $tan(x)=\beta$

What is the value of $\alpha$ in function of $\beta$ ?

Possible Answers:

$\alpha=\frac{2\beta}{1-\beta^2}$

$\alpha=\frac{2\beta^2}{1-\beta^2}$

$\alpha=\frac{2\beta}{1-2\beta^2}$

$\alpha=\frac{2\beta}{2-\beta^2}$

$\alpha=\frac{2\beta}{3-2\beta^2}$

Correct answer:

$\alpha=\frac{2\beta}{1-\beta^2}$

Explanation:

Using trigonometric identities we know that :

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

This gives :

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

We also know that $\alpha =tan(2x)$

This gives :

$\alpha=\frac{2\beta}{1-\beta^2}$

Report an Error

View Pre-Calculus Tutors

Nicholas
Certified Tutor

Fordham University, Bachelors, Business Administration and Management. Boston University, Master of Arts, Multimedia.

View Pre-Calculus Tutors

Joshua
Certified Tutor

Cornerstone University, Bachelor in Arts, Elementary School Teaching.

View Pre-Calculus Tutors

Dylan
Certified Tutor

University of California-Berkeley, Bachelor of Science, Cognitive Science.

All Precalculus Resources

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

Popular Subjects

GMAT Tutors in Chicago, Biology Tutors in San Diego, Computer Science Tutors in New York City, ACT Tutors in Phoenix, MCAT Tutors in Houston, Physics Tutors in Phoenix, GMAT Tutors in San Francisco-Bay Area, GMAT Tutors in Miami, GMAT Tutors in Philadelphia, SAT Tutors in Washington DC

Popular Courses & Classes

ISEE Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Atlanta, SAT Courses & Classes in Houston, Spanish Courses & Classes in Chicago, GRE Courses & Classes in Denver, MCAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Los Angeles, GRE Courses & Classes in Boston, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Washington DC

Popular Test Prep

LSAT Test Prep in Boston, MCAT Test Prep in Miami, MCAT Test Prep in Philadelphia, MCAT Test Prep in Houston, LSAT Test Prep in San Diego, SSAT Test Prep in New York City, SSAT Test Prep in Los Angeles, ISEE Test Prep in San Francisco-Bay Area, ISEE Test Prep in Phoenix, SSAT Test Prep in Seattle

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 : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1502 : Pre Calculus

Example Question #1502 : Pre Calculus

Example Question #1503 : Pre Calculus

Example Question #23 : Fundamental Trigonometric Identities

Example Question #144 : Trigonometric Functions

Example Question #146 : Trigonometric Functions

Example Question #31 : Fundamental Trigonometric Identities

Example Question #31 : Fundamental Trigonometric Identities

Example Question #31 : Fundamental Trigonometric Identities

Example Question #147 : Trigonometric Functions

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: