Answer: C) \( \frac{-120}{169} \)
Solution:
We know the identity:
\[
\sin^2 x + \cos^2 x = 1
\]
Let \( \sin x + \cos x = \frac{5}{13} \). Squaring both sides:
\[
(\sin x + \cos x)^2 = \frac{25}{169}
\]
Expanding the square:
\[
\sin^2 x + \cos^2 x + 2\sin x \cos x = \frac{25}{169}
\]
Substitute \( \sin^2 x + \cos^2 x = 1 \):
\[
1 + 2\sin x \cos x = \frac{25}{169}
\]
Simplify:
\[
2\sin x \cos x = \frac{25}{169} - 1 = \frac{25}{169} - \frac{169}{169} = \frac{-144}{169}
\]
Since \( \sin 2x = 2\sin x \cos x \), we have:
\[
\sin 2x = \frac{-144}{169}
\]
Final Answer:
- \( \sin 2x = \frac{-144}{169} \), so the correct option is \( \text{C} \).
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