Call Now to Set Up Tutoring:

(888) 888-0446

Trigonometry : Identities of Halved Angles

Study concepts, example questions & explanations for Trigonometry

Create An Account Create Tests & Flashcards

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 :Trigonometry

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

Possible Answers:

$\frac{1}{5}$

$\frac{24}{25}$

$-\frac{12}{25}$

$-\frac{24}{25}$

$-\frac{3}{5}$

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 #1 :Identities Of Halved Angles

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

Possible Answers:

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

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

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

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

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

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 #53 :Trigonometric Identities

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

Possible Answers:

$1-cos(4\theta)$

$1+cos(2\theta)$

$1-cos(2\theta)$

$1-sin(4\theta)$

$1+cos(4\theta)$

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 #54 :Trigonometric Identities

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 ,因此……

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

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

Report an Error

Example Question #55 :Trigonometric Identities

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

Possible Answers:

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

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

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

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

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

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 #56 :Trigonometric Identities

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

Possible Answers:

$\frac{1}{2}$

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

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

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

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 #51 :Trigonometric Identities

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

Possible Answers:

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

$\frac{1}{2}$

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

$\frac{\sqrt3}{2}$

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

View Trigonometry Tutors

Michael
Certified Tutor

University of Richmond, Bachelor of Science, Mathematics. Central Connecticut State University, Master of Arts, Mathematics.

View Trigonometry Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View Trigonometry Tutors

Rodrigo
Certified Tutor

University of California-Davis, Bachelor of Science, Applied Mathematics.

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Trigonometry Tutoring in Top Cities:

Trigonometry Tutors in Top Cities:

Popular Courses & Classes

GMAT Courses & Classes in Atlanta,ACT Courses & Classes in Seattle,SAT Courses & Classes in Phoenix,MCAT Courses & Classes in Seattle,ACT Courses & Classes in Miami,ACT Courses & Classes in New York City,LSAT Courses & Classes in Houston,SSAT Courses & Classes in Los Angeles,SSAT Courses & Classes in Seattle,MCAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

LSAT Test Prep in Houston,SSAT Test Prep in New York City,MCAT Test Prep in Houston,SSAT Test Prep in San Diego,LSAT Test Prep in New York City,GMAT Test Prep in Chicago,SSAT Test Prep in Dallas Fort Worth,MCAT Test Prep in Dallas Fort Worth,LSAT Test Prep in Seattle,SAT Test Prep in Houston

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 : Identities of Halved Angles

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #51 :Trigonometry

Example Question #1 :Identities Of Halved Angles

Example Question #53 :Trigonometric Identities

Example Question #54 :Trigonometric Identities

Example Question #55 :Trigonometric Identities

Example Question #56 :Trigonometric Identities

Example Question #51 :Trigonometric Identities

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: