Q. 38 find the exact value of each of ... [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. 38 (page 486)

find the exact value of each of the following under the given conditions:
$\cos 伪 = \frac{1}{2}, - \frac{蟺}{2} < 伪 < 0; \sin 尾 = \frac{1}{3}, 0 < 尾 < \frac{蟺}{2}$
role="math" localid="1647314573285" $a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$

Short Answer

Expert verified

The values are,

$a) \sin (伪 + 尾) = \frac{1 - 2 \sqrt{6}}{6} b) \cos (伪 + 尾) = \frac{2 \sqrt{2} + \sqrt{3}}{6} c) \sin (伪 - 尾) = \frac{- 2 \sqrt{6} - 1}{6} d) \tan (伪 - 尾) = \frac{- 2 \sqrt{6} - 1}{2 \sqrt{2} - \sqrt{3}}$

Step by step solution

Given information

We are given $\cos 伪 = \frac{1}{2}, - \frac{蟺}{2} < 伪 < 0; \sin 尾 = \frac{1}{3}, 0 < 尾 < \frac{蟺}{2}$

we find sinα

We have,

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

We find cosβ

We have,

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

Find sin(α+β)

We have,

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

Find cos(α+β)

We have,

$co s (伪 + 尾) = \cos 伪 \cos 尾 - \sin 伪 \sin 尾 co s (伪 + 尾) = (\frac{1}{2}) (\frac{2 \sqrt{2}}{3}) - (- \frac{\sqrt{3}}{2}) (\frac{1}{3}) co s (伪 + 尾) = \frac{2 \sqrt{2}}{6} + \frac{\sqrt{3}}{6} co s (伪 + 尾) = \frac{2 \sqrt{2} + \sqrt{3}}{6}$

Find sin(α-β)

We have,

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

Find tanα ,tanβ

We have,

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

Similarly

$\tan 尾 = \frac{\sin 尾}{\cos 尾} \tan 尾 = \frac{\frac{1}{3}}{\frac{2 \sqrt{2}}{3}} \tan 尾 = \frac{1}{2 \sqrt{2}}$

Now,

$\tan (伪 - 尾) = \frac{\tan 伪 - \tan 尾}{1 + \tan 伪 \tan 尾} \tan (伪 - 尾) = \frac{- \sqrt{3} - \frac{1}{2 \sqrt{2}}}{1 + (- \sqrt{3}) (\frac{1}{2 \sqrt{2}})} \tan (伪 - 尾) = \frac{- 2 \sqrt{6} - 1}{2 \sqrt{2} - \sqrt{3}}$

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91影视

find the exact value of each of the following under the given conditions:
$\cos 伪 = \frac{1}{2}, - \frac{蟺}{2} < 伪 < 0; \sin 尾 = \frac{1}{3}, 0 < 尾 < \frac{蟺}{2}$
role="math" localid="1647314573285" $a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$

Short Answer

Step by step solution

Given information

we find sinα

We find cosβ

Find sin(α+β)

Find cos(α+β)

Find sin(α-β)

Find tanα ,tanβ

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

find the exact value of each of the following under the given conditions: cos伪=12,-蟺2<伪<0;sin尾=13,0<尾<蟺2role="math" localid="1647314573285" a)sin(伪+尾)b)cos(伪+尾)c)sin(伪-尾)d)tan(伪-尾)

Short Answer

Step by step solution

Given information

we find sinα

We find cosβ

Find sin(α+β)

Find cos(α+β)

Find sin(α-β)

Find tanα ,tanβ

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Calculus

Statistics

Decision Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

find the exact value of each of the following under the given conditions:
$\cos 伪 = \frac{1}{2}, - \frac{蟺}{2} < 伪 < 0; \sin 尾 = \frac{1}{3}, 0 < 尾 < \frac{蟺}{2}$
role="math" localid="1647314573285" $a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$