Certainly! Let's break this problem down step-by-step.
1. Given Information:
- [tex]\(\sin \theta = \frac{12}{13}\)[/tex]
- [tex]\(\theta\)[/tex] is in Quadrant II
2. Characteristics of Quadrant II:
- In Quadrant II, sine ([tex]\(\sin \theta\)[/tex]) is positive.
- Cosine ([tex]\(\cos \theta\)[/tex]) is negative.
3. Finding [tex]\(\cos \theta\)[/tex]:
- We use the Pythagorean identity: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
- Plug in the given value of [tex]\(\sin \theta\)[/tex]:
[tex]\[
\left(\frac{12}{13}\right)^2 + \cos^2 \theta = 1
\][/tex]
- Simplify the equation:
[tex]\[
\frac{144}{169} + \cos^2 \theta = 1
\][/tex]
- Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}
\][/tex]
- Since [tex]\(\cos \theta\)[/tex] is negative in Quadrant II:
[tex]\[
\cos \theta = -\sqrt{\frac{25}{169}} = -\frac{5}{13}
\][/tex]
4. Finding [tex]\(\cos 2\theta\)[/tex]:
- Using the double-angle formula for cosine: [tex]\(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\)[/tex].
- Substitute [tex]\(\cos \theta = -\frac{5}{13}\)[/tex] and [tex]\(\sin \theta = \frac{12}{13}\)[/tex] into the formula:
[tex]\[
\cos 2\theta = \left(-\frac{5}{13}\right)^2 - \left(\frac{12}{13}\right)^2
\][/tex]
[tex]\[
\cos 2\theta = \frac{25}{169} - \frac{144}{169}
\][/tex]
[tex]\[
\cos 2\theta = \frac{25 - 144}{169} = \frac{-119}{169} = -\frac{119}{169}
\][/tex]
Given the numerical results, we find that:
- [tex]\(\cos \theta = -0.38461538461538447\)[/tex]
- [tex]\(\cos 2\theta = -0.7041420118343198\)[/tex]
Therefore, completing the statement:
If [tex]\(\sin \theta = \frac{12}{13}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant II, [tex]\(\cos 2\theta = -0.7041420118343198\)[/tex] and [tex]\(\cos \theta = -0.38461538461538447\)[/tex].
