Q. 11 Use the information given about ... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 11 (page 495)

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

Short Answer

Expert verified

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

part (b). $\cos 2 胃 = \frac{1}{3}$

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

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

part (e). $\tan 2 胃 = - 2 \sqrt{2}$

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

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,

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

We know that,

$\sin 2 胃 = 2 \sin 胃 \cos 胃 \sin 2 胃 = 2 脳 \frac{1}{\sqrt{3}} 脳 \frac{- \sqrt{6}}{3} \sin 2 胃 = - \frac{2 \sqrt{2}}{3}$

Part (b) Step 1. value of cos2θ

We know that,

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

part (c) Step 1. Value of sinθ2

We know that,

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

Part (d) step 4. Value of cosθ2

We know that,

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

Part (e) step 1. Value of tan2θ

We have:

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

We know that,

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

Part (f) step 1. value of tanθ2

We have:

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

We know that,

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

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91影视

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

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

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91影视

Use the information given about the angle 0<胃<2蟺, to find the exact value of cos胃=-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

Geometry

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Calculus

Theoretical and Mathematical Physics

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Applied Mathematics

Study anywhere. Anytime. Across all devices.

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