Given that $\sin \left(\theta \right)$ = $-\frac{12}{13}$, and that $\theta$ is in the third quadrant, what is the value of $\cos \left(\theta \right)$?

• A. $-\frac{12}{13}$
• B. $\frac{5}{13}$
• C. $-\frac{5}{13}$
• D. $\frac{12}{13}$

Answer: (C)

Verified step by step guidance

1. Recall the Pythagorean identity: \(\sin^{2}(\theta) + \cos^{2}(\theta) = 1\).
2. Substitute the given value of \(\sin(\theta) = -\frac{12}{13}\) into the identity: \(\left(-\frac{12}{13}\right)^{2} + \cos^{2}(\theta) = 1\).
3. Calculate \(\sin^{2}(\theta)\): \(\left(-\frac{12}{13}\right)^{2} = \frac{144}{169}\), so the equation becomes \(\frac{144}{169} + \cos^{2}(\theta) = 1\).
4. Solve for \(\cos^{2}(\theta)\): \(\cos^{2}(\theta) = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}\).
5. Take the square root of both sides to find \(\cos(\theta)\): \(\cos(\theta) = \pm \frac{5}{13}\). Since \(\theta\) is in the third quadrant, where cosine is negative, choose \(\cos(\theta) = -\frac{5}{13}\).