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Sure, let's solve this step-by-step.
Given:
[tex]\[ \sin \alpha = \frac{21}{29}, \quad \cos \beta = \frac{-9}{41} \][/tex]
First, we need to find the missing trigonometric values for [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
### 1. Finding [tex]\(\cos \alpha\)[/tex]
Using the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Therefore,
[tex]\[ \left(\frac{21}{29}\right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{441}{841} + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \cos^2 \alpha = 1 - \frac{441}{841} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{841 - 441}{841} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{400}{841} \][/tex]
[tex]\[ \cos \alpha = \sqrt{\frac{400}{841}} \][/tex]
[tex]\[ \cos \alpha = \frac{20}{29} \][/tex]
Since [tex]\(\alpha\)[/tex] is in the first quadrant, [tex]\(\cos \alpha\)[/tex] is positive:
[tex]\[ \cos \alpha = \frac{20}{29} \][/tex]
### 2. Finding [tex]\(\sin \beta\)[/tex]
Using the Pythagorean identity:
[tex]\[ \cos^2 \beta + \sin^2 \beta = 1 \][/tex]
Therefore,
[tex]\[ \left(\frac{-9}{41}\right)^2 + \sin^2 \beta = 1 \][/tex]
[tex]\[ \frac{81}{1681} + \sin^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta = 1 - \frac{81}{1681} \][/tex]
[tex]\[ \sin^2 \beta = \frac{1681 - 81}{1681} \][/tex]
[tex]\[ \sin^2 \beta = \frac{1600}{1681} \][/tex]
[tex]\[ \sin \beta = \sqrt{\frac{1600}{1681}} \][/tex]
[tex]\[ \sin \beta = \frac{40}{41} \][/tex]
Since [tex]\(\beta\)[/tex] is in the first quadrant, [tex]\(\sin \beta\)[/tex] is positive:
[tex]\[ \sin \beta = \frac{40}{41} \][/tex]
### a) [tex]\( \sin (\alpha + \beta) \)[/tex]
Using the angle addition formula for sine:
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
Substitute the known values:
[tex]\[ \sin (\alpha + \beta) = \frac{21}{29} \cdot \frac{-9}{41} + \frac{20}{29} \cdot \frac{40}{41} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{21 \cdot -9}{29 \cdot 41} + \frac{20 \cdot 40}{29 \cdot 41} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-189}{1189} + \frac{800}{1189} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-189 + 800}{1189} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{611}{1189} \][/tex]
### b) [tex]\( \cos (\alpha + \beta) \)[/tex]
Using the angle addition formula for cosine:
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
Substitute the known values:
[tex]\[ \cos (\alpha + \beta) = \frac{20}{29} \cdot \frac{-9}{41} - \frac{21}{29} \cdot \frac{40}{41} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{20 \cdot -9}{29 \cdot 41} - \frac{21 \cdot 40}{29 \cdot 41} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-180}{1189} - \frac{840}{1189} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-1020}{1189} \][/tex]
### c) [tex]\( \tan (\alpha - \beta) \)[/tex]
First, we need to find [tex]\(\tan \alpha\)[/tex] and [tex]\(\tan \beta\)[/tex]:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{21}{29}}{\frac{20}{29}} = \frac{21}{20} \][/tex]
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{40}{41}}{\frac{-9}{41}} = -\frac{40}{9} \][/tex]
Using the angle subtraction formula for tangent:
[tex]\[ \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \][/tex]
Substitute the known values:
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21}{20} - \left( -\frac{40}{9} \right)}{1 + \left( \frac{21}{20} \cdot -\frac{40}{9} \right)} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21}{20} + \frac{40}{9}}{1 - \frac{21 \cdot 40}{20 \cdot 9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21 \cdot 9 + 40 \cdot 20}{20 \cdot 9}}{1 - \frac{21 \cdot 40}{20 \cdot 9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{189 + 800}{180}}{1 - \frac{840}{180}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{1 - \frac{42}{9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{1 - \frac{14}{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{\frac{-11}{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{989}{180} \cdot \frac{-3}{11} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{989 \cdot -3}{180 \cdot 11} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-2967}{1980} \][/tex]
[tex]\[ \tan (\alpha - \beta) = -\frac{2967}{1980} \][/tex]
[tex]\[ \tan (\alpha - \beta) = -\frac{989}{660} \][/tex]
Thus, the answers are:
a) [tex]\(\sin (\alpha + \beta) = \frac{611}{1189}\)[/tex]
b) [tex]\(\cos (\alpha + \beta) = \frac{-1020}{1189}\)[/tex]
c) [tex]\(\tan (\alpha - \beta) = -\frac{989}{660}\)[/tex]
Given:
[tex]\[ \sin \alpha = \frac{21}{29}, \quad \cos \beta = \frac{-9}{41} \][/tex]
First, we need to find the missing trigonometric values for [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
### 1. Finding [tex]\(\cos \alpha\)[/tex]
Using the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Therefore,
[tex]\[ \left(\frac{21}{29}\right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{441}{841} + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \cos^2 \alpha = 1 - \frac{441}{841} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{841 - 441}{841} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{400}{841} \][/tex]
[tex]\[ \cos \alpha = \sqrt{\frac{400}{841}} \][/tex]
[tex]\[ \cos \alpha = \frac{20}{29} \][/tex]
Since [tex]\(\alpha\)[/tex] is in the first quadrant, [tex]\(\cos \alpha\)[/tex] is positive:
[tex]\[ \cos \alpha = \frac{20}{29} \][/tex]
### 2. Finding [tex]\(\sin \beta\)[/tex]
Using the Pythagorean identity:
[tex]\[ \cos^2 \beta + \sin^2 \beta = 1 \][/tex]
Therefore,
[tex]\[ \left(\frac{-9}{41}\right)^2 + \sin^2 \beta = 1 \][/tex]
[tex]\[ \frac{81}{1681} + \sin^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta = 1 - \frac{81}{1681} \][/tex]
[tex]\[ \sin^2 \beta = \frac{1681 - 81}{1681} \][/tex]
[tex]\[ \sin^2 \beta = \frac{1600}{1681} \][/tex]
[tex]\[ \sin \beta = \sqrt{\frac{1600}{1681}} \][/tex]
[tex]\[ \sin \beta = \frac{40}{41} \][/tex]
Since [tex]\(\beta\)[/tex] is in the first quadrant, [tex]\(\sin \beta\)[/tex] is positive:
[tex]\[ \sin \beta = \frac{40}{41} \][/tex]
### a) [tex]\( \sin (\alpha + \beta) \)[/tex]
Using the angle addition formula for sine:
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
Substitute the known values:
[tex]\[ \sin (\alpha + \beta) = \frac{21}{29} \cdot \frac{-9}{41} + \frac{20}{29} \cdot \frac{40}{41} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{21 \cdot -9}{29 \cdot 41} + \frac{20 \cdot 40}{29 \cdot 41} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-189}{1189} + \frac{800}{1189} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-189 + 800}{1189} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{611}{1189} \][/tex]
### b) [tex]\( \cos (\alpha + \beta) \)[/tex]
Using the angle addition formula for cosine:
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
Substitute the known values:
[tex]\[ \cos (\alpha + \beta) = \frac{20}{29} \cdot \frac{-9}{41} - \frac{21}{29} \cdot \frac{40}{41} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{20 \cdot -9}{29 \cdot 41} - \frac{21 \cdot 40}{29 \cdot 41} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-180}{1189} - \frac{840}{1189} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-1020}{1189} \][/tex]
### c) [tex]\( \tan (\alpha - \beta) \)[/tex]
First, we need to find [tex]\(\tan \alpha\)[/tex] and [tex]\(\tan \beta\)[/tex]:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{21}{29}}{\frac{20}{29}} = \frac{21}{20} \][/tex]
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{40}{41}}{\frac{-9}{41}} = -\frac{40}{9} \][/tex]
Using the angle subtraction formula for tangent:
[tex]\[ \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \][/tex]
Substitute the known values:
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21}{20} - \left( -\frac{40}{9} \right)}{1 + \left( \frac{21}{20} \cdot -\frac{40}{9} \right)} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21}{20} + \frac{40}{9}}{1 - \frac{21 \cdot 40}{20 \cdot 9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{21 \cdot 9 + 40 \cdot 20}{20 \cdot 9}}{1 - \frac{21 \cdot 40}{20 \cdot 9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{189 + 800}{180}}{1 - \frac{840}{180}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{1 - \frac{42}{9}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{1 - \frac{14}{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\frac{989}{180}}{\frac{-11}{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{989}{180} \cdot \frac{-3}{11} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{989 \cdot -3}{180 \cdot 11} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-2967}{1980} \][/tex]
[tex]\[ \tan (\alpha - \beta) = -\frac{2967}{1980} \][/tex]
[tex]\[ \tan (\alpha - \beta) = -\frac{989}{660} \][/tex]
Thus, the answers are:
a) [tex]\(\sin (\alpha + \beta) = \frac{611}{1189}\)[/tex]
b) [tex]\(\cos (\alpha + \beta) = \frac{-1020}{1189}\)[/tex]
c) [tex]\(\tan (\alpha - \beta) = -\frac{989}{660}\)[/tex]
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