Answer :
Given that \(\cos \theta = \frac{24}{25}\) and \(\theta\) lies in quadrant IV, we need to find the exact values for \(\sin 2\theta\), \(\cos 2\theta\), and \(\tan 2\theta\).
### a. \(\sin 2\theta\)
We start with the double angle formula for sine:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
First, we find \(\sin \theta\). Since \(\theta\) is in quadrant IV, \(\sin \theta\) will be negative. Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
we know \(\cos \theta = \frac{24}{25}\), so:
[tex]\[ \sin^2 \theta + \left(\frac{24}{25}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{576}{625} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{576}{625} \][/tex]
[tex]\[ \sin^2 \theta = \frac{625}{625} - \frac{576}{625} \][/tex]
[tex]\[ \sin^2 \theta = \frac{49}{625} \][/tex]
So, \(\sin \theta = -\frac{7}{25}\) (since \(\theta\) is in quadrant IV).
Now, substituting \(\sin \theta\) and \(\cos \theta\) into the double angle formula:
[tex]\[ \sin 2\theta = 2 \left(-\frac{7}{25}\right) \left(\frac{24}{25}\right) \][/tex]
[tex]\[ \sin 2\theta = 2 \left(-\frac{168}{625}\right) \][/tex]
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
Thus:
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
### b. \(\cos 2\theta\)
We use the double angle formula for cosine:
[tex]\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \][/tex]
Substituting the known values:
[tex]\[ \cos 2\theta = \left(\frac{24}{25}\right)^2 - \left(-\frac{7}{25}\right)^2 \][/tex]
[tex]\[ \cos 2\theta = \frac{576}{625} - \frac{49}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{576 - 49}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
Thus:
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
### c. \(\tan 2\theta\)
We use the double angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \][/tex]
Substituting the values we found:
[tex]\[ \tan 2\theta = \frac{-\frac{336}{625}}{\frac{527}{625}} \][/tex]
[tex]\[ \tan 2\theta = \frac{-336}{527} \][/tex]
Thus:
[tex]\[ \tan 2\theta = -\frac{336}{527} \][/tex]
### Summary
Given the information:
[tex]\[ \cos \theta = \frac{24}{25}, \theta \text{ lies in quadrant IV} \][/tex]
we have found:
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
[tex]\[ \tan 2\theta = -\frac{336}{527} \][/tex]
### a. \(\sin 2\theta\)
We start with the double angle formula for sine:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
First, we find \(\sin \theta\). Since \(\theta\) is in quadrant IV, \(\sin \theta\) will be negative. Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
we know \(\cos \theta = \frac{24}{25}\), so:
[tex]\[ \sin^2 \theta + \left(\frac{24}{25}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{576}{625} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{576}{625} \][/tex]
[tex]\[ \sin^2 \theta = \frac{625}{625} - \frac{576}{625} \][/tex]
[tex]\[ \sin^2 \theta = \frac{49}{625} \][/tex]
So, \(\sin \theta = -\frac{7}{25}\) (since \(\theta\) is in quadrant IV).
Now, substituting \(\sin \theta\) and \(\cos \theta\) into the double angle formula:
[tex]\[ \sin 2\theta = 2 \left(-\frac{7}{25}\right) \left(\frac{24}{25}\right) \][/tex]
[tex]\[ \sin 2\theta = 2 \left(-\frac{168}{625}\right) \][/tex]
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
Thus:
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
### b. \(\cos 2\theta\)
We use the double angle formula for cosine:
[tex]\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \][/tex]
Substituting the known values:
[tex]\[ \cos 2\theta = \left(\frac{24}{25}\right)^2 - \left(-\frac{7}{25}\right)^2 \][/tex]
[tex]\[ \cos 2\theta = \frac{576}{625} - \frac{49}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{576 - 49}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
Thus:
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
### c. \(\tan 2\theta\)
We use the double angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \][/tex]
Substituting the values we found:
[tex]\[ \tan 2\theta = \frac{-\frac{336}{625}}{\frac{527}{625}} \][/tex]
[tex]\[ \tan 2\theta = \frac{-336}{527} \][/tex]
Thus:
[tex]\[ \tan 2\theta = -\frac{336}{527} \][/tex]
### Summary
Given the information:
[tex]\[ \cos \theta = \frac{24}{25}, \theta \text{ lies in quadrant IV} \][/tex]
we have found:
[tex]\[ \sin 2\theta = -\frac{336}{625} \][/tex]
[tex]\[ \cos 2\theta = \frac{527}{625} \][/tex]
[tex]\[ \tan 2\theta = -\frac{336}{527} \][/tex]
