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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 10 (page 495)

Use the information given about the angle , to find the exact value of
$\tan 胃 = \frac{1}{2}, 蟺 < 胃 < \frac{3 蟺}{2}$

Short Answer

Expert verified

Part (a). $\sin 2 胃 = \frac{4}{5}$

Part (b). $\cos 2 胃 = \frac{3}{5}$

Part (c). $\sin \frac{胃}{2} = \sqrt{\frac{2 + \sqrt{5}}{2 \sqrt{5}}}$

Part (d). $\cos \frac{胃}{2} = = \sqrt{\frac{\sqrt{5} - 2}{2 \sqrt{5}}}$

part (e). $\tan 2 胃 = \frac{4}{3}$

Part (f). $\tan \frac{胃}{2} = \frac{\sqrt{2 + \sqrt{5}}}{\sqrt{\sqrt{5} - 2}}$

Step by step solution

01

Part (a) Step 1. Value of sin2θ

Angle is in the third quadrant.

So, sine and cosine are negative.

$\sin 胃 = - \frac{1}{\sqrt{5}} \cos 胃 = - \frac{2}{\sqrt{5}}$

we know that,

$\sin 2 胃 = 2 \sin 胃 \cos 胃 = 2 脳 (- \frac{1}{\sqrt{5}}) (- \frac{2}{\sqrt{5}}) = \frac{4}{5}$

02

Part(b) step 1. Value of cos2θ

we know that,

$\cos 2 胃 = \cos^{2} 胃 - \sin^{2} 胃 = {(- \frac{2}{\sqrt{5}})}^{2} - (- \frac{1}{\sqrt{5}}) = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}$

03

Part (c) Step 1. Value of sinθ2

We know that,

$\sin \frac{胃}{2} = \sqrt{\frac{1 - \cos 胃}{2}} = \sqrt{\frac{1 - (- \frac{2}{\sqrt{5}})}{2}} = \sqrt{\frac{2 + \sqrt{5}}{2 \sqrt{5}}}$

04

Part (d) Step 1. Value of cosθ2

We know that,

$\cos 2 胃 = \sqrt{\frac{1 + \cos 胃}{2}} = \sqrt{\frac{1 + (- \frac{2}{\sqrt{5}})}{2}} = \sqrt{\frac{\sqrt{5} - 2}{2 \sqrt{5}}}$

05

Part (e) Step 1. Value of tan2θ

We have:

$\sin 2 胃 = \frac{4}{5} \cos 2 胃 = \frac{3}{5}$

we know that,

$\tan 2 胃 = \frac{\sin 2 胃}{\cos 2 胃} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$

06

Part (f) Step 1. Value of tanθ2

We have:

$\sin \frac{胃}{2} = \sqrt{\frac{2 + \sqrt{5}}{2 \sqrt{5}}} \cos \frac{胃}{2} = \sqrt{\frac{\sqrt{5} - 2}{2 \sqrt{5}}}$

We know that,

role="math" localid="1646544830476"

\tan \frac{胃}{2} = \frac{\sin \frac{胃}{2}}{\cos \frac{胃}{2}} \tan \frac{胃}{2} = \frac{\sqrt{\frac{2 + \sqrt{5}}{2 \sqrt{5}}}}{\sqrt{\frac{\sqrt{5} - 2}{2 \sqrt{5}}}} \tan \frac{胃}{2} = \frac{\sqrt{2 + \sqrt{5}}}{\sqrt{\sqrt{5} - 2}}

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Most popular questions from this chapter

Establish each identity.

$\frac{1 + \sin 胃 + \cos 胃}{1 + \sin 胃 - \cos 胃} = \frac{1 + \cos 胃}{\sin 胃}$

solve each equation on the interval

$\sqrt{3} \sin 胃 + \cos 胃 = 1$

express each product as a sum containing only sines or only cosines

$\sin \frac{3 蠎}{2} \cos \frac{蠎}{2}$

Show that $\sin^{- 1} (v) + \cos^{- 1} (v) = \frac{蟺}{2}$

If $x = 2 \tan 胃$ , express $\sin (2 胃)$ as a function ofx.

See all solutions

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