Chapter 7: Q. 11 (page 495)

Use the information given about the angle $0 < θ < 2 π$ , to find the exact value of
$\cos θ = \frac{- \sqrt{6}}{3}$

Short Answer

Expert verified

Part (a). $\sin 2 θ = - \frac{2 \sqrt{2}}{3}$

part (b). $\cos 2 θ = \frac{1}{3}$

part (c). $\sin \frac{θ}{2} = \sqrt{\frac{3 + \sqrt{6}}{6}}$

part(d). $\cos \frac{θ}{2} = \sqrt{\frac{3 - \sqrt{6}}{6}}$

part (e). $\tan 2 θ = - 2 \sqrt{2}$

part (f). $\tan \frac{θ}{2} = \sqrt{\frac{3 - \sqrt{6}}{3 + \sqrt{6}}}$

Step by step solution

Part (a) Step 1. Value of sin2θ

Angle is in the second quadrant.

So, sine is positive and cosine is negative.

So,

$\sin θ = \sqrt{1 - \cos^{2} θ} \sin θ = \sqrt{1 - {(- \frac{\sqrt{6}}{3})}^{2}} \sin θ = \sqrt{1 - \frac{6}{9}} \sin θ = \frac{1}{\sqrt{3}}$

We know that,

$\sin 2 θ = 2 \sin θ \cos θ \sin 2 θ = 2 \times \frac{1}{\sqrt{3}} \times \frac{- \sqrt{6}}{3} \sin 2 θ = - \frac{2 \sqrt{2}}{3}$

Part (b) Step 1. value of cos2θ

We know that,

$\cos 2 θ = \cos^{2} θ - \sin^{2} θ = {(\frac{- \sqrt{6}}{3})}^{2} - {(\frac{1}{\sqrt{3}})}^{2} = \frac{6}{9} - \frac{1}{3} = \frac{1}{3}$

part (c) Step 1. Value of sinθ2

We know that,

$\sin \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{2}} = \sqrt{\frac{1 - (- \frac{\sqrt{6}}{3})}{2}} = \sqrt{\frac{3 + \sqrt{6}}{6}}$

Part (d) step 4. Value of cosθ2

We know that,

$\cos \frac{θ}{2} = \sqrt{\frac{1 + \cos θ}{2}} = \frac{\sqrt{1 + (- \frac{\sqrt{6}}{3})}}{2} = \sqrt{\frac{3 - \sqrt{6}}{6}}$

Part (e) step 1. Value of tan2θ

We have:

$\sin 2 θ = - \frac{2 \sqrt{2}}{3} \cos 2 θ = \frac{1}{3}$

We know that,

role="math" localid="1646561005169" $\tan 2 θ = \frac{\sin 2 θ}{\cos 2 θ} = \frac{- \frac{2 \sqrt{2}}{3}}{\frac{1}{3}} = - 2 \sqrt{2}$

Part (f) step 1. value of tanθ2

We have:

$\sin \frac{θ}{2} = \sqrt{\frac{3 + \sqrt{6}}{6}} \cos \frac{θ}{2} = \sqrt{\frac{3 - \sqrt{6}}{6}}$

We know that,

$\tan \frac{θ}{2} = \frac{\sqrt{\frac{3 + \sqrt{6}}{6}}}{\sqrt{\frac{3 - \sqrt{6}}{6}}} = \sqrt{\frac{3 - \sqrt{6}}{3 + \sqrt{6}}}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Use the information given about the angle $0 < θ < 2 π$ , to find the exact value of
$\cos θ = \frac{- \sqrt{6}}{3}$

Short Answer

Step by step solution

Part (a) Step 1. Value of sin2θ

Part (b) Step 1. value of cos2θ

part (c) Step 1. Value of sinθ2

Part (d) step 4. Value of cosθ2

Part (e) step 1. Value of tan2θ

Part (f) step 1. value of tanθ2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Applied Mathematics

Pure Maths

Mechanics Maths

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

Use the information given about the angle 0<θ<2π, to find the exact value of cosθ=-63

Short Answer

Step by step solution

Part (a) Step 1. Value of sin2θ

Part (b) Step 1. value of cos2θ

part (c) Step 1. Value of sinθ2

Part (d) step 4. Value of cosθ2

Part (e) step 1. Value of tan2θ

Part (f) step 1. value of tanθ2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Applied Mathematics

Pure Maths

Mechanics Maths

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

Use the information given about the angle $0 < θ < 2 π$ , to find the exact value of
$\cos θ = \frac{- \sqrt{6}}{3}$