Certainly, let’s break down the solution step-by-step.
We are given:
[tex]\[ \cos \alpha = \frac{-12}{13} \][/tex]
[tex]\[ \cot \beta = \frac{24}{7} \][/tex]
[tex]\(\alpha\)[/tex] is in Quadrant II and [tex]\(\beta\)[/tex] is in Quadrant III.
We need to find [tex]\(\sin(\alpha+\beta)\)[/tex], [tex]\(\cos(\alpha+\beta)\)[/tex], and [tex]\(\tan(\alpha+\beta)\)[/tex].
### Step 1: Find [tex]\(\sin \alpha\)[/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\sin \alpha > 0\)[/tex].
We know that:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \cos^2 \alpha \][/tex]
Substitute the given value of [tex]\(\cos \alpha\)[/tex]:
[tex]\[ \sin^2 \alpha = 1 - \left(\frac{-12}{13}\right)^2 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{144}{169} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{169 - 144}{169} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{25}{169} \][/tex]
[tex]\[ \sin \alpha = \frac{5}{13} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{25}{169}} \][/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\sin \alpha\)[/tex] is positive:
[tex]\[ \sin \alpha = \frac{5}{13} \approx 0.3846 \][/tex]
### Step 2: Find [tex]\(\sin \beta\)[/tex] and [tex]\(\cos \beta\)[/tex]
Given [tex]\(\cot \beta = \frac{24}{7}\)[/tex], which indicates [tex]\(\beta\)[/tex] is in Quadrant III. In this quadrant, both [tex]\(\sin \beta < 0\)[/tex] and [tex]\(\cos \beta < 0\)[/tex].
[tex]\[
\tan \beta = \frac{1}{\cot \beta} = \frac{7}{24}
\][/tex]
[tex]\[
\tan \beta = \frac{\sin \beta}{\cos \beta}
\][/tex]
Using the identity [tex]\(\tan^2 \beta + 1 = \frac{1}{\cos^2 \beta}\)[/tex]:
[tex]\[
\tan^2 \beta = \left(\frac{7}{24}\right)^2 = \frac{49}{576}
\][/tex]
[tex]\[
1 + \tan^2 \beta = 1 + \frac{49}{576} = \frac{576 + 49}{576} = \frac{625}{576}
\][/tex]
[tex]\[
\cos^2 \beta = \frac{1}{1 + \tan^2 \beta} = \frac{576}{625}
\][/tex]
[tex]\[
\cos \beta = \sqrt{\frac{576}{625}} = \frac{24}{25}
\][/tex]
Since [tex]\(\cos \beta\)[/tex] is negative in Quadrant III,
[tex]\[
\cos \beta = -0.96
\][/tex]
Now to find [tex]\(\sin \beta\)[/tex]:
[tex]\[
\tan \beta = \frac{\sin \beta}{\cos \beta} \Rightarrow \sin \beta = \tan \beta \cdot \cos \beta
\][/tex]
[tex]\[
\sin \beta = \frac{7}{24} \cdot -0.96 = -0.28
\][/tex]
### Step 3: Finding [tex]\(\sin (\alpha + \beta)\)[/tex], [tex]\(\cos (\alpha + \beta)\)[/tex], and [tex]\(\tan (\alpha + \beta)\)[/tex]
Using the sum identities,
[tex]\[
\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta
\][/tex]
[tex]\[
\sin (\alpha + \beta) = \left(\frac{5}{13}\right) \left(-0.96\right) + \left(-\frac{12}{13}\right)\left(-0.28\right)
\][/tex]
[tex]\[
\sin (\alpha + \beta) = -0.627 + 0.269 = -0.6277
\][/tex]
[tex]\[
\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
\][/tex]
[tex]\[
\cos (\alpha + \beta) = \left(-\frac{12}{13}\right) \left(-0.96\right) - \left(\frac{5}{13}\right) \left(-0.28\right)
\][/tex]
[tex]\[
\cos (\alpha + \beta) = 0.8884 + 0.1077 = 0.7784
\][/tex]
Finally, we find [tex]\(\tan (\alpha + \beta)\)[/tex] using the formula:
[tex]\[
\tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)}
\][/tex]
[tex]\[
\tan (\alpha + \beta) = \frac{-0.6277}{0.7784} \approx -0.8063
\][/tex]
Thus, the values are:
[tex]\[
\sin (\alpha + \beta) = -0.6277
\][/tex]
[tex]\[
\cos (\alpha + \beta) = 0.7784
\][/tex]
[tex]\[
\tan (\alpha + \beta) = -0.8063
\][/tex]
These are the required trigonometric values.
