To find [tex]\(\sin(2A)\)[/tex] given that [tex]\(\sin A = -\frac{12}{13}\)[/tex] and [tex]\(A\)[/tex] is in the third quadrant, we follow these steps:
1. Recognize that in the third quadrant, both sine and cosine values are negative. Given [tex]\(\sin A = -\frac{12}{13}\)[/tex], we need to find [tex]\(\cos A\)[/tex].
2. Use the Pythagorean identity:
[tex]\[
\sin^2(A) + \cos^2(A) = 1
\][/tex]
3. Substitute the value of [tex]\(\sin A\)[/tex] and solve for [tex]\(\cos^2(A)\)[/tex]:
[tex]\[
\left(-\frac{12}{13}\right)^2 + \cos^2(A) = 1
\][/tex]
[tex]\[
\frac{144}{169} + \cos^2(A) = 1
\][/tex]
[tex]\[
\cos^2(A) = 1 - \frac{144}{169}
\][/tex]
[tex]\[
\cos^2(A) = \frac{169}{169} - \frac{144}{169}
\][/tex]
[tex]\[
\cos^2(A) = \frac{25}{169}
\][/tex]
4. Take the square root of both sides to find [tex]\(\cos A\)[/tex]. Since [tex]\(A\)[/tex] is in the third quadrant, [tex]\(\cos A\)[/tex] is negative:
[tex]\[
\cos A = -\sqrt{\frac{25}{169}}
\][/tex]
[tex]\[
\cos A = -\frac{5}{13}
\][/tex]
5. Now, use the double-angle identity for sine:
[tex]\[
\sin(2A) = 2 \sin(A) \cos(A)
\][/tex]
6. Substitute [tex]\(\sin A = -\frac{12}{13}\)[/tex] and [tex]\(\cos A = -\frac{5}{13}\)[/tex]:
[tex]\[
\sin(2A) = 2 \left(-\frac{12}{13}\right) \left(-\frac{5}{13}\right)
\][/tex]
[tex]\[
\sin(2A) = 2 \left(\frac{60}{169}\right)
\][/tex]
[tex]\[
\sin(2A) = \frac{120}{169}
\][/tex]
Thus, the value of [tex]\(\sin(2A)\)[/tex] is approximately:
[tex]\[
\sin(2A) \approx 0.710059
\][/tex]
Hence, [tex]\(\sin(2A) = \frac{120}{169} \approx 0.710059\)[/tex].
