Find the exact value of each of the following under the given conditions:

[tex]\[
\tan \alpha = -\frac{7}{24}, \quad \alpha \text{ lies in quadrant II}, \quad \cos \beta = \frac{3}{4}, \quad \beta \text{ lies in quadrant I}
\][/tex]

a. \(\sin (\alpha + \beta) = \) \(\square\)
- Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions.

b. \(\cos (\alpha + \beta) = \) \(\square\)
- Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions.

c. \(\tan (\alpha + \beta) = \) \(\square\)
- Simplify your answer. Type an exact answer, using radicals as needed. Rationalize all denominators. Use integers or fractions.



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]