Q. 35 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. 35 (page 486)

find the exact value of each of the following under the given conditions: $\tan 伪 = - \frac{4}{3}, \frac{蟺}{2} < 伪 < 蟺; c o s 尾 = \frac{1}{2}, 0 < 尾 < \frac{蟺}{2}$
$a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$

Short Answer

Expert verified

The values are,

$a) \sin (伪 + 尾) = \frac{4 - 3 \sqrt{3}}{10} b) \cos (伪 + 尾) = \frac{- 3 - 4 \sqrt{3}}{10} c) \sin (伪 - 尾) = \frac{4 + 3 \sqrt{3}}{10} d) \tan (伪 - 尾) = \frac{4 + \sqrt{3}}{39}$

Step by step solution

Given information

We are given $\tan 伪 = - \frac{4}{3}, \frac{蟺}{2} < 伪 < 蟺; c o s 尾 = \frac{1}{2}, 0 < 尾 < \frac{蟺}{2}$

we find cosα,sinα

We have,

$\cos 伪 = \sqrt{\frac{1}{\tan^{2} 伪 + 1}} \cos 伪 = \sqrt{\frac{1}{{(- \frac{4}{3})}^{2} + 1}} \cos 伪 = - \frac{3}{5} a s 伪 i s i n s e c o n d q u a d r a n t$

Similarly

$\sin 伪 = \sqrt{1 - \cos^{2} 伪} \sin 伪 = \sqrt{1 - {(- \frac{3}{5})}^{2}} \sin 伪 = \sqrt{1 - \frac{9}{25}} \sin 伪 = \frac{4}{5}$

Find sinβ

We have,

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

Find sin(α+β)

We have,

$\sin (伪 + 尾) = \sin 伪 \cos 尾 + \cos 伪 \sin 尾 \sin (伪 + 尾) = (\frac{4}{5}) (\frac{1}{2}) + (- \frac{3}{5}) (\frac{\sqrt{3}}{2}) \sin (伪 + 尾) = \frac{4}{10} - \frac{3 \sqrt{3}}{10} \sin (伪 + 尾) = \frac{4 - 3 \sqrt{3}}{10}$

Find cos(α+β)

We have,

$co s (伪 + 尾) = \cos 伪 \cos 尾 - \sin 伪 \sin 尾 co s (伪 + 尾) = (- \frac{3}{5}) (\frac{1}{2}) - (\frac{4}{5}) (\frac{\sqrt{3}}{2}) co s (伪 + 尾) = - \frac{3}{10} - \frac{4 \sqrt{3}}{10} co s (伪 + 尾) = \frac{- 4 \sqrt{3} - 3}{10}$

Find sin(α-β)

We have,

$\sin (伪 - 尾) = \sin 伪 \cos 尾 - \cos 伪 \sin 尾 \sin (伪 - 尾) = (\frac{4}{5}) (\frac{1}{2}) - (- \frac{3}{5}) (\frac{\sqrt{3}}{2}) \sin (伪 - 尾) = \frac{4}{10} + \frac{3 \sqrt{3}}{10} \sin (伪 - 尾) = \frac{4 + 3 \sqrt{3}}{10}$

Find tan(α-β)

We have,

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

now we substitute the values in the formula

localid="1647311051057" $\tan (伪 - 尾) = \frac{\tan 伪 - \tan 尾}{1 + \tan 伪 \tan 尾} \tan (伪 - 尾) = \frac{- \frac{4}{3} - \sqrt{3}}{1 + (- \frac{4}{3}) (\sqrt{3})} \tan (伪 - 尾) = \frac{4 + \sqrt{3}}{39}$

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91影视

find the exact value of each of the following under the given conditions: $\tan 伪 = - \frac{4}{3}, \frac{蟺}{2} < 伪 < 蟺; c o s 尾 = \frac{1}{2}, 0 < 尾 < \frac{蟺}{2}$
$a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$

Short Answer

Step by step solution

Given information

we find cosα,sinα

Find sinβ

Find sin(α+β)

Find cos(α+β)

Find sin(α-β)

Find tan(α-β)

One App. One Place for Learning.

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91影视

find the exact value of each of the following under the given conditions:tan伪=-43,蟺2<伪<蟺;cos尾=12,0<尾<蟺2a)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 cosα,sinα

Find sinβ

Find sin(α+β)

Find cos(α+β)

Find sin(α-β)

Find tan(α-β)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Calculus

Mechanics Maths

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Statistics

Study anywhere. Anytime. Across all devices.

find the exact value of each of the following under the given conditions: $\tan 伪 = - \frac{4}{3}, \frac{蟺}{2} < 伪 < 蟺; c o s 尾 = \frac{1}{2}, 0 < 尾 < \frac{蟺}{2}$
$a) \sin (伪 + 尾) b) \cos (伪 + 尾) c) \sin (伪 - 尾) d) \tan (伪 - 尾)$