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 : Identities Of Halved Angles

Find $\sin 2\theta$ if $\cos \theta = \frac{4}{5}$ and $\frac{3\pi}{2}<\theta<2\pi$ .

Possible Answers:

$\frac{24}{25}$

$-\frac{3}{5}$

$-\frac{24}{25}$

$\frac{1}{5}$

$-\frac{12}{25}$

Correct answer:

$-\frac{24}{25}$

Explanation:

The double-angle identity for sine is written as

$\sin 2\theta= 2\sin\theta \cos\theta$

and we know that

$\cos\theta = \frac{4}{5}$

Using $x^{2}+y^{2}=r^{2}$ , we see that $(4)^{2} + (y)^{2} = (5)^{2}$ , which gives us

$y=\pm3$

Since we know $\theta$ is between $\frac{3\pi}{2}$ and $2\pi$ , sin $\theta$ is negative, so y= -3 . Thus,

$\sin \theta = -\frac{3}{5}$ .

Finally, substituting into our double-angle identity, we get

$\sin 2\theta = 2 \sin\theta \cos\theta = 2 \cdot\bigg(-\frac{3}{5}\bigg)\cdot\frac{4}{5} = -\frac{24}{25}$

Report an Error

Example Question #51 : Trigonometry

Find the exact value of $\sin 15^{\circ}$ using an appropriate half-angle identity.

Possible Answers:

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

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

$\frac{1-\sqrt{3}}{4}$

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

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

Correct answer:

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

Explanation:

The half-angle identity for sine is:

$\sin \bigg(\frac{x}{2}\bigg) = \pm\sqrt\frac{1-\cos x}{2}$

If our half-angle is $15^{\circ}$ , then our full angle is $30^{\circ}$ . Thus,

$\sin 15^{\circ}= \pm\sqrt\frac{1-\cos30^{\circ}}{2}$

The exact value of $\cos 30^{\circ}$ is expressed as $\frac{\sqrt{3}}{2}$ , so we have

$\sin 15^{\circ} = \pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$

Simplify under the outer radical and we get

$\sin 15^{\circ} = \pm\sqrt\frac{2-\sqrt{3}}{4}$

Now simplify the denominator and get

$\sin 15^{\circ} = \pm\frac{\sqrt{2-\sqrt{3}}}{2}$

Since $15^{\circ}$ is in the first quadrant, we know sin is positive. So,

$\sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2}$

Report an Error

Example Question #3 : Identities Of Halved Angles

Which of the following best represents $2cos^2(2\theta)$ ?

Possible Answers:

$1-cos(2\theta)$

$1+cos(4\theta)$

$1-cos(4\theta)$

$1+cos(2\theta)$

$1-sin(4\theta)$

Correct answer:

$1+cos(4\theta)$

Explanation:

Write the half angle identity for cosine.

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

Replace theta with two theta.

$\theta=2\theta$

Therefore:

$2cos^2(2\theta)= 2\left[\frac{1+cos(2 \times 2\theta)}{2}\right] = 1+cos(4\theta)$

Report an Error

Example Question #52 : Trigonometry

What is the amplitude of 8cos^2(x) ?

Possible Answers:

Correct answer:

Explanation:

The key here is to use the half-angle identity for cos^2(x) to convert it and make it much easier to work with.

$acos^2(x) = \frac{a}{2}[1 + cos(x)]$

In this case, a = 8 , so therefore...

$8cos^2(x) = \frac{8}{2}[1 + cos(x)] = 4 + 4cos(x)$

Consequently, 8cos^2(x) has an amplitude of .

Report an Error

Example Question #51 : Trigonometry

If $cos(45^{\circ})=\frac{\sqrt{2}}{2}$ , then calculate $cos(22.5^{\circ})$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Because $22.5^{\circ}=\frac{45^{\circ}}{2}$ , we can use the half-angle formula for cosines to determine $cos(22.5^{\circ})$ .

In general,

$cos(\frac{\Theta}{2})= \sqrt{\frac{1}{2}(1+cos\Theta)}$

for $0^{\circ}\leq\Theta<360^{\circ}$ .

For this problem,

$cos(22.5^{\circ})= \sqrt{\frac{1}{2}(1+cos(45^{\circ}))}$

$= \sqrt{\frac{1}{2}(1+\frac{\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{2}(\frac{2+\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{4}(2+\sqrt{2})}$

$=\frac{1}{2}\sqrt{(2+\sqrt{2})}$

Hence,

$cos(22.5^{\circ})=\frac{1}{2}\sqrt{2+\sqrt{2}}$

Report an Error

Example Question #5 : Identities Of Halved Angles

What is $\cos(\frac{5\pi}{12})$ ?

Possible Answers:

$\frac{\sqrt{4-\sqrt{6}}}{2}$

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

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

$\frac{1}{2}$

Correct answer:

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

Explanation:

Let $x=\frac{5\pi}{6}$ ; then

$\cos(\frac{5\pi}{12})=\cos(\frac{x}{2})$ .

We'll use the half-angle formula to evaluate this expression.

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

Now we'll substitute $\frac{5\pi}{6}$ for .

$\begin{align*} \cos(\frac{5\pi}{12})=\cos(\frac{\frac{5\pi}{6}}{2})&=\pm\sqrt{\frac{1+\cos(\frac{5\pi}{6})}{2}}\\ &=\pm\sqrt{\frac{1+\frac{-\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{2-\sqrt{3}}{2}}\\ &=\pm\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right) \end{align*}$

$\frac{5\pi}{6}$ is in the first quadrant, so $\cos(\frac{5\pi}{6})$ is positive. So

$\cos(\frac{5\pi}{12})=\frac{\sqrt{2-\sqrt{3}}}{2}$ .

Report an Error

Example Question #1 : Identities Of Halved Angles

What is $\tan(15)$ , given that $\sin(30)$ and $\cos(30)$ are well defined values?

Possible Answers:

$\frac{\frac{\sqrt3}{2}}{\frac{3}{2}}$

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

$\frac{\sqrt3}{2}$

$\frac{1}{2}$

Correct answer:

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

Explanation:

Using the half angle formula for tangent,

$\tan(\frac{\theta}{2})=\frac{\sin(\theta)}{(1+\cos(\theta))}$ ,

we plug in 30 for $\theta$ .

We also know from the unit circle that $\sin(30)$ is 1/2 and $\cos(30)$ is $\sqrt3/2$ .

Plug all values into the equation, and you will get the correct answer.

$\\ \tan(\frac{30}{2})=\frac{\sin(30)}{(1+\cos(30))} \\ \\ \\ \tan(15)=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$

Report an Error

Example Question #51 : Trigonometric Identities

Using trigonometric identities prove whether the following is valid:

$\sin^{2}A + \cot^{2}A = \frac{2\sin^{4}A +2\cos^{2}A}{1 - \cos 2A}$

Possible Answers:

Only in the range of: $0 < A < 2\pi$

Only in the range of: $0 < A < \frac{\pi}{2}$

Uncertain

False

True

Correct answer:

True

Explanation:

We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following:

$\sin^{2}A + \cot^{2}A = \frac{2\left(\sin^{4}A + \cos^{2}A\right)}{1-\cos 2A}$

Next we note the power reducing formula for sine so we can extract the necessary components as follows:

$\sin^{2}A + \cot^{2}A = \frac{2}{1-\cos 2A}\left(\sin^{4}A + \cos^{2}A\right)$

The power reducing formula must be inverted giving:

$\sin^{2}A + \cot^{2}A = \frac{1}{\sin^{2}A}\left(\sin^{4}A + \cos^{2}A\right)$

Now we can distribute and reduce:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \frac{\cos^{2}A}{\sin^{2}A}$

Finally recalling the basic identity for the cotangent:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \cot^{2}A$

This proves the equivalence.

Report an Error

Example Question #52 : Trigonometric Identities

Use the power reducing formulas for trigonometric functions to reduce and simplify the following equation:

$\sin^{2}A \cos^{2} A$

Possible Answers:

$1 + \cos^{2}A$

$1 + \cos 2A$

$\frac{1 + \cos^{2}2A}{2}$

$\frac{3 + \cos 4A}{8}$

$\frac{1 + 2\cos A}{2}$

Correct answer:

$\frac{3 + \cos 4A}{8}$

Explanation:

The power reducing formulas for both sine and cosine differ in only the operation in the numerator. Applying the power reducing formulas here we get:

$\frac{1-\cos 2A}{2}\times \frac{1 + \cos 2A}{2}$

Multiplying the binomials in the numerator and multiplying the denominators:
$\frac{1-\cos 2A + \cos 2A - \cos^{2} 2A}{4}$

Reducing the numerator:

$\frac{1+\cos^{2}2A}{4}$

We again use the power reducing formula for cosine as follows:

$\frac{1 + \frac{1 + \cos 4A}{2}}{4}$

Combining the numerator by determining a common denominator:

$\frac{\frac{3 + \cos 4A}{2}}{4}$

Now simply reducing the double fraction:

$\frac{3 + \cos 4A}{8}$

Report an Error

Example Question #1 : Identities Of Squared Trigonometric Functions

Using trigonometric identities, determine whether the following is valid:

$\sin^{2} A \cos^{2} A = \frac{1-\cos^{4}A-2\cos^{2}A\sin^{2}A + \sin^{4}A}{4}$

Possible Answers:

Only valid in the range of: $0 < A < \frac{\pi}{2}$

False

True

Only valid in the range of: $0 < A < 2\pi$

Uncertain

Correct answer:

False

Explanation:

In this case we choose to work with the side that appears to be simpler, the left hand side. We begin by using the power reducing formulas:

$\frac{1-\cos 2A}{2}\frac{1+\cos 2A}{2} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Next we perform the multiplication on the numerator:

$\frac{1-\cos^{2}2A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

The next step we take is to remove the double angle, since there is no double angle in the alleged solution:

$\frac{1 - \left(\cos^{2}A -\sin^{2}A\right )^{2}}{4}=\frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Finally we multiply the binomials in the numerator on the left hand side to determine if the equivalence holds:

$\frac{1-\cos^{4}A +2\cos^{2}A\sin^{2}A-\sin^{4}A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

We see that the equivalence does not hold.

Report an Error

View Trigonometry Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Trigonometry Tutors

Vandana
Certified Tutor

Jawaharlal Nehru Technological University, Bachelor of Technology, Civil Engineering. Florida International University, Maste...

View Trigonometry Tutors

Vicky
Certified Tutor

Clemson University, Doctor of Science, Civil Engineering.

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

SAT Courses & Classes in Seattle, Spanish Courses & Classes in Seattle, SAT Courses & Classes in Houston, LSAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Philadelphia, GRE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Washington DC, MCAT Courses & Classes in Chicago, GMAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GMAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Miami, GMAT Test Prep in Philadelphia, SAT Test Prep in Philadelphia, ISEE Test Prep in Seattle, ISEE Test Prep in Dallas Fort Worth, SSAT Test Prep in Washington DC, GMAT Test Prep in San Diego, GRE Test Prep in Los Angeles, ISEE Test Prep in San Diego

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 : Identities Of Halved Angles

Example Question #51 : Trigonometry

Example Question #3 : Identities Of Halved Angles

Example Question #52 : Trigonometry

Example Question #51 : Trigonometry

Example Question #5 : Identities Of Halved Angles

Example Question #1 : Identities Of Halved Angles

Example Question #51 : Trigonometric Identities

Example Question #52 : Trigonometric Identities

Example Question #1 : Identities Of Squared Trigonometric Functions

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: