Answer :
Let's solve this step by step according to the given conditions. We have:
[tex]\[ \tan \alpha = -\frac{7}{24}, \quad \alpha \text{ lies in quadrant II} \][/tex]
[tex]\[ \cos \beta = \frac{3}{4}, \quad \beta \text{ lies in quadrant I} \][/tex]
### Step 1: Find \(\sin \alpha\) and \(\cos \alpha\)
Since \(\alpha\) lies in quadrant II, \(\cos \alpha\) will be negative and \(\sin \alpha\) will be positive.
To find \(\cos \alpha\), we use the identity:
[tex]\[ \tan^2 \alpha + 1 = \sec^2 \alpha \][/tex]
Given \(\tan \alpha = -\frac{7}{24}\):
[tex]\[ \left(-\frac{7}{24}\right)^2 + 1 = \sec^2 \alpha \Rightarrow \frac{49}{576} + 1 = \sec^2 \alpha \Rightarrow \frac{49}{576} + \frac{576}{576} = \sec^2 \alpha \Rightarrow \sec^2 \alpha = \frac{625}{576} \][/tex]
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \sec \alpha = \frac{25}{24} \Rightarrow \cos \alpha = \frac{24}{25} \][/tex]
As \(\alpha\) is in quadrant II, \(\cos \alpha\) is negative:
[tex]\[ \cos \alpha = -\frac{24}{25} \][/tex]
Now, using \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\):
[tex]\[ \tan \alpha = -\frac{7}{24} = \frac{\sin \alpha}{-\frac{24}{25}} \Rightarrow \sin \alpha = \frac{7}{25} \][/tex]
So, we have:
[tex]\[ \sin \alpha = \frac{7}{25} \][/tex]
### Step 2: Find \(\sin \beta\)
Given \(\cos \beta = \frac{3}{4}\) and \(\beta \text{ lies in quadrant I}\), where both \(\sin \beta\) and \(\cos \beta\) are positive, we use:
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta + \left(\frac{3}{4}\right)^2 = 1 \Rightarrow \sin^2 \beta + \frac{9}{16} = 1 \Rightarrow \sin^2 \beta = \frac{7}{16} \][/tex]
Taking the positive root:
[tex]\[ \sin \beta = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \][/tex]
### Step 3: Calculate \(\sin (\alpha + \beta)\), \(\cos (\alpha + \beta)\), and \(\tan (\alpha + \beta)\)
#### a. \(\sin (\alpha + \beta)\)
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
Substitute the values:
[tex]\[ \sin (\alpha + \beta) = \left(\frac{7}{25}\right) \left(\frac{3}{4}\right) + \left(-\frac{24}{25}\right) \left(\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{21}{100} - \frac{24\sqrt{7}}{100} = \frac{21 - 24\sqrt{7}}{100} \][/tex]
#### b. \(\cos (\alpha + \beta)\)
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
Substitute the values:
[tex]\[ \cos (\alpha + \beta) = \left(-\frac{24}{25}\right) \left(\frac{3}{4}\right) - \left(\frac{7}{25}\right) \left(\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{72}{100} - \frac{7\sqrt{7}}{100} = \frac{-72 - 7\sqrt{7}}{100} \][/tex]
#### c. \(\tan (\alpha + \beta)\)
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3} \][/tex]
Use the identity:
[tex]\[ \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \][/tex]
Substitute the values:
[tex]\[ \tan (\alpha + \beta) = \frac{-\frac{7}{24} + \frac{\sqrt{7}}{3}}{1 - \left(-\frac{7}{24}\right) \left(\frac{\sqrt{7}}{3}\right)} \][/tex]
[tex]\[ \tan (\alpha + \beta) = \frac{-\frac{7}{24} + \frac{8\sqrt{7}}{24}}{1 + \frac{7\sqrt{7}}{72}} = \frac{\frac{-7 + 8\sqrt{7}}{24}}{\frac{72 + 7\sqrt{7}}{72}} = \frac{-7 + 8\sqrt{7}}{24 + \frac{72}{72}7\sqrt{7}} = \frac{-7 + 8\sqrt{7}}{31\sqrt{33}} } \][/tex]
So the exact values are:
[tex]\[ \sin (\alpha + \beta) = \boxed{\frac{21 - 24\sqrt{7}}{100}}, \quad \cos (\alpha + \beta) = \boxed{\frac{-72 - 7\sqrt{7}}{100}}, \quad \tan (\alpha + \beta) = \boxed{0.469} \][/tex]
[tex]\[ \tan \alpha = -\frac{7}{24}, \quad \alpha \text{ lies in quadrant II} \][/tex]
[tex]\[ \cos \beta = \frac{3}{4}, \quad \beta \text{ lies in quadrant I} \][/tex]
### Step 1: Find \(\sin \alpha\) and \(\cos \alpha\)
Since \(\alpha\) lies in quadrant II, \(\cos \alpha\) will be negative and \(\sin \alpha\) will be positive.
To find \(\cos \alpha\), we use the identity:
[tex]\[ \tan^2 \alpha + 1 = \sec^2 \alpha \][/tex]
Given \(\tan \alpha = -\frac{7}{24}\):
[tex]\[ \left(-\frac{7}{24}\right)^2 + 1 = \sec^2 \alpha \Rightarrow \frac{49}{576} + 1 = \sec^2 \alpha \Rightarrow \frac{49}{576} + \frac{576}{576} = \sec^2 \alpha \Rightarrow \sec^2 \alpha = \frac{625}{576} \][/tex]
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \sec \alpha = \frac{25}{24} \Rightarrow \cos \alpha = \frac{24}{25} \][/tex]
As \(\alpha\) is in quadrant II, \(\cos \alpha\) is negative:
[tex]\[ \cos \alpha = -\frac{24}{25} \][/tex]
Now, using \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\):
[tex]\[ \tan \alpha = -\frac{7}{24} = \frac{\sin \alpha}{-\frac{24}{25}} \Rightarrow \sin \alpha = \frac{7}{25} \][/tex]
So, we have:
[tex]\[ \sin \alpha = \frac{7}{25} \][/tex]
### Step 2: Find \(\sin \beta\)
Given \(\cos \beta = \frac{3}{4}\) and \(\beta \text{ lies in quadrant I}\), where both \(\sin \beta\) and \(\cos \beta\) are positive, we use:
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \sin^2 \beta + \left(\frac{3}{4}\right)^2 = 1 \Rightarrow \sin^2 \beta + \frac{9}{16} = 1 \Rightarrow \sin^2 \beta = \frac{7}{16} \][/tex]
Taking the positive root:
[tex]\[ \sin \beta = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \][/tex]
### Step 3: Calculate \(\sin (\alpha + \beta)\), \(\cos (\alpha + \beta)\), and \(\tan (\alpha + \beta)\)
#### a. \(\sin (\alpha + \beta)\)
[tex]\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \][/tex]
Substitute the values:
[tex]\[ \sin (\alpha + \beta) = \left(\frac{7}{25}\right) \left(\frac{3}{4}\right) + \left(-\frac{24}{25}\right) \left(\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{21}{100} - \frac{24\sqrt{7}}{100} = \frac{21 - 24\sqrt{7}}{100} \][/tex]
#### b. \(\cos (\alpha + \beta)\)
[tex]\[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \][/tex]
Substitute the values:
[tex]\[ \cos (\alpha + \beta) = \left(-\frac{24}{25}\right) \left(\frac{3}{4}\right) - \left(\frac{7}{25}\right) \left(\frac{\sqrt{7}}{4}\right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = -\frac{72}{100} - \frac{7\sqrt{7}}{100} = \frac{-72 - 7\sqrt{7}}{100} \][/tex]
#### c. \(\tan (\alpha + \beta)\)
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3} \][/tex]
Use the identity:
[tex]\[ \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \][/tex]
Substitute the values:
[tex]\[ \tan (\alpha + \beta) = \frac{-\frac{7}{24} + \frac{\sqrt{7}}{3}}{1 - \left(-\frac{7}{24}\right) \left(\frac{\sqrt{7}}{3}\right)} \][/tex]
[tex]\[ \tan (\alpha + \beta) = \frac{-\frac{7}{24} + \frac{8\sqrt{7}}{24}}{1 + \frac{7\sqrt{7}}{72}} = \frac{\frac{-7 + 8\sqrt{7}}{24}}{\frac{72 + 7\sqrt{7}}{72}} = \frac{-7 + 8\sqrt{7}}{24 + \frac{72}{72}7\sqrt{7}} = \frac{-7 + 8\sqrt{7}}{31\sqrt{33}} } \][/tex]
So the exact values are:
[tex]\[ \sin (\alpha + \beta) = \boxed{\frac{21 - 24\sqrt{7}}{100}}, \quad \cos (\alpha + \beta) = \boxed{\frac{-72 - 7\sqrt{7}}{100}}, \quad \tan (\alpha + \beta) = \boxed{0.469} \][/tex]
