To determine [tex]\( \sin(2x) \)[/tex] given the conditions [tex]\( \cos(x) = -\frac{1}{4} \)[/tex] and [tex]\( \tan(x) < 0 \)[/tex], follow these steps:
1. Determine the quadrant where [tex]\(x\)[/tex] lies:
- [tex]\( \cos(x) < 0 \)[/tex], so [tex]\(x\)[/tex] must be in either the second or third quadrant.
- [tex]\( \tan(x) < 0 \)[/tex], meaning [tex]\( \sin(x) \)[/tex] and [tex]\( \cos(x) \)[/tex] have opposite signs. Therefore, [tex]\(x\)[/tex] is in the second quadrant.
2. Find [tex]\( \sin(x) \)[/tex] using the Pythagorean identity:
- Recall the identity: [tex]\( \sin^2(x) + \cos^2(x) = 1 \)[/tex]
- Substitute [tex]\( \cos(x) \)[/tex]:
[tex]\[
\sin^2(x) + \left(-\frac{1}{4}\right)^2 = 1
\][/tex]
[tex]\[
\sin^2(x) + \frac{1}{16} = 1
\][/tex]
- Solve for [tex]\( \sin^2(x) \)[/tex]:
[tex]\[
\sin^2(x) = 1 - \frac{1}{16}
\][/tex]
[tex]\[
\sin^2(x) = \frac{16}{16} - \frac{1}{16}
\][/tex]
[tex]\[
\sin^2(x) = \frac{15}{16}
\][/tex]
[tex]\[
\sin(x) = \sqrt{\frac{15}{16}} \text{ or } \sin(x) = -\sqrt{\frac{15}{16}}
\][/tex]
- Since [tex]\(x\)[/tex] is in the second quadrant where sine is positive:
[tex]\[
\sin(x) = \frac{\sqrt{15}}{4}
\][/tex]
3. Calculate [tex]\( \sin(2x) \)[/tex] using the double-angle formula:
- The formula is: [tex]\( \sin(2x) = 2 \sin(x) \cos(x) \)[/tex]
- Substitute [tex]\( \sin(x) \)[/tex] and [tex]\( \cos(x) \)[/tex]:
[tex]\[
\sin(2x) = 2 \cdot \frac{\sqrt{15}}{4} \cdot -\frac{1}{4}
\][/tex]
- Simplify the expression:
[tex]\[
\sin(2x) = 2 \cdot \frac{\sqrt{15}}{4} \cdot -\frac{1}{4} = 2 \cdot \frac{\sqrt{15} \cdot -1}{16}
\][/tex]
[tex]\[
\sin(2x) = \frac{-2 \sqrt{15}}{16}
\][/tex]
[tex]\[
\sin(2x) = - \frac{\sqrt{15}}{8}
\][/tex]
Thus, the exact value of [tex]\( \sin(2x) \)[/tex] is [tex]\( - \frac{\sqrt{15}}{8} \)[/tex]. The corresponding choice is:
[tex]\[
\boxed{-\frac{\sqrt{15}}{8}}
\][/tex]
