Answer :
Let's solve for each of these values step by step under the given conditions.
Given:
- [tex]\(\tan \alpha = -\frac{5}{12}\)[/tex] with [tex]\(\frac{\pi}{2} < \alpha < \pi\)[/tex]
- [tex]\(\cos \beta = \frac{1}{2}\)[/tex] with [tex]\(0 < \beta < \frac{\pi}{2}\)[/tex]
### Step-by-Step Solutions:
#### 1. Determine [tex]\(\sin \alpha\)[/tex] and [tex]\(\cos \alpha\)[/tex]:
[tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{5}{12}\)[/tex]
Given [tex]\(\frac{\pi}{2} < \alpha < \pi\)[/tex], [tex]\(\sin \alpha\)[/tex] is positive and [tex]\(\cos \alpha\)[/tex] is negative.
Using the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
We relate the tangent to the sides of a right triangle where:
[tex]\(\text{opposite} = 5\)[/tex], [tex]\(\text{adjacent} = -12\)[/tex].
The hypotenuse [tex]\( = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)[/tex].
Thus:
[tex]\[ \sin \alpha = \frac{5}{13} \][/tex]
[tex]\[ \cos \alpha = -\frac{12}{13} \][/tex]
#### 2. Determine [tex]\(\sin \beta\)[/tex]:
Given that [tex]\(\cos \beta = \frac{1}{2}\)[/tex], find [tex]\(\sin \beta\)[/tex].
Since [tex]\(0 < \beta < \frac{\pi}{2}\)[/tex], [tex]\(\sin \beta\)[/tex] is positive.
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta + \left(\frac{1}{2}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \beta + \frac{1}{4} = 1 \][/tex]
[tex]\[ \sin^2 \beta = \frac{3}{4} \][/tex]
[tex]\[ \sin \beta = \frac{\sqrt{3}}{2} \][/tex]
#### 3. Find the required values:
(a) [tex]\(\sin (\alpha + \beta)\)[/tex]:
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
[tex]\[ \sin (\alpha + \beta) = \left(\frac{5}{13}\right) \left(\frac{1}{2}\right) + \left(-\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{5}{26} - \frac{12\sqrt{3}}{26} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{5 - 12\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(-0.6071003727240972\)[/tex].
(b) [tex]\(\cos (\alpha + \beta)\)[/tex]:
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
[tex]\[ \cos (\alpha + \beta) = \left(-\frac{12}{13}\right) \left(\frac{1}{2}\right) - \left(\frac{5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{12}{26} - \frac{5\sqrt{3}}{26} \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{12 + 5\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(-0.7946251553017072\)[/tex].
(c) [tex]\(\sin (\alpha - \beta)\)[/tex]:
[tex]\[ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \][/tex]
[tex]\[ \sin (\alpha - \beta) = \left(\frac{5}{13}\right) \left(\frac{1}{2}\right) - \left(-\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \sin (\alpha - \beta) = \frac{5}{26} + \frac{12\sqrt{3}}{26} \][/tex]
[tex]\[ \sin (\alpha - \beta) = \frac{5 + 12\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(0.9917157573394818\)[/tex].
(d) [tex]\(\tan (\alpha - \beta)\)[/tex]:
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5}{12} - \sqrt{3}}{1 + \left(-\frac{5}{12}\right) \sqrt{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5}{12} - \frac{12\sqrt{3}}{12}}{1 - \frac{5\sqrt{3}}{12}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5 + 12\sqrt{3}}{12}}{\frac{12 - 5\sqrt{3}}{12}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-(5 + 12\sqrt{3})}{12 - 5\sqrt{3}} \][/tex]
The numerical result is [tex]\(-7.720530238828117\)[/tex].
These are the calculated values for each of the requirements.
Given:
- [tex]\(\tan \alpha = -\frac{5}{12}\)[/tex] with [tex]\(\frac{\pi}{2} < \alpha < \pi\)[/tex]
- [tex]\(\cos \beta = \frac{1}{2}\)[/tex] with [tex]\(0 < \beta < \frac{\pi}{2}\)[/tex]
### Step-by-Step Solutions:
#### 1. Determine [tex]\(\sin \alpha\)[/tex] and [tex]\(\cos \alpha\)[/tex]:
[tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{5}{12}\)[/tex]
Given [tex]\(\frac{\pi}{2} < \alpha < \pi\)[/tex], [tex]\(\sin \alpha\)[/tex] is positive and [tex]\(\cos \alpha\)[/tex] is negative.
Using the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
We relate the tangent to the sides of a right triangle where:
[tex]\(\text{opposite} = 5\)[/tex], [tex]\(\text{adjacent} = -12\)[/tex].
The hypotenuse [tex]\( = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)[/tex].
Thus:
[tex]\[ \sin \alpha = \frac{5}{13} \][/tex]
[tex]\[ \cos \alpha = -\frac{12}{13} \][/tex]
#### 2. Determine [tex]\(\sin \beta\)[/tex]:
Given that [tex]\(\cos \beta = \frac{1}{2}\)[/tex], find [tex]\(\sin \beta\)[/tex].
Since [tex]\(0 < \beta < \frac{\pi}{2}\)[/tex], [tex]\(\sin \beta\)[/tex] is positive.
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta + \left(\frac{1}{2}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \beta + \frac{1}{4} = 1 \][/tex]
[tex]\[ \sin^2 \beta = \frac{3}{4} \][/tex]
[tex]\[ \sin \beta = \frac{\sqrt{3}}{2} \][/tex]
#### 3. Find the required values:
(a) [tex]\(\sin (\alpha + \beta)\)[/tex]:
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
[tex]\[ \sin (\alpha + \beta) = \left(\frac{5}{13}\right) \left(\frac{1}{2}\right) + \left(-\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{5}{26} - \frac{12\sqrt{3}}{26} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{5 - 12\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(-0.6071003727240972\)[/tex].
(b) [tex]\(\cos (\alpha + \beta)\)[/tex]:
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
[tex]\[ \cos (\alpha + \beta) = \left(-\frac{12}{13}\right) \left(\frac{1}{2}\right) - \left(\frac{5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{12}{26} - \frac{5\sqrt{3}}{26} \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{12 + 5\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(-0.7946251553017072\)[/tex].
(c) [tex]\(\sin (\alpha - \beta)\)[/tex]:
[tex]\[ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \][/tex]
[tex]\[ \sin (\alpha - \beta) = \left(\frac{5}{13}\right) \left(\frac{1}{2}\right) - \left(-\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ \sin (\alpha - \beta) = \frac{5}{26} + \frac{12\sqrt{3}}{26} \][/tex]
[tex]\[ \sin (\alpha - \beta) = \frac{5 + 12\sqrt{3}}{26} \][/tex]
The numerical result is [tex]\(0.9917157573394818\)[/tex].
(d) [tex]\(\tan (\alpha - \beta)\)[/tex]:
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5}{12} - \sqrt{3}}{1 + \left(-\frac{5}{12}\right) \sqrt{3}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5}{12} - \frac{12\sqrt{3}}{12}}{1 - \frac{5\sqrt{3}}{12}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-\frac{5 + 12\sqrt{3}}{12}}{\frac{12 - 5\sqrt{3}}{12}} \][/tex]
[tex]\[ \tan (\alpha - \beta) = \frac{-(5 + 12\sqrt{3})}{12 - 5\sqrt{3}} \][/tex]
The numerical result is [tex]\(-7.720530238828117\)[/tex].
These are the calculated values for each of the requirements.
