To solve this problem, let's go through the steps.

Step 1: Understand the Given Information

We know:

1. $\cot x = -\frac{1}{2}$
2. $x$ is in the 4th quadrant.

In the 4th quadrant:

• $\cos x$ is positive.
• $\sin x$ is negative.

Step 2: Relate $\cot x$ to $\sin x$ and $\cos x$

Since $\cot x = \frac{\cos x}{\sin x}$, we can set up the equation: $\cot x = -\frac{1}{2} = \frac{\cos x}{\sin x}.$

Let’s assume $\cos x = k$ and $\sin x = -2k$ (keeping the signs consistent with quadrant IV).

Step 3: Use the Pythagorean Identity

Since $\sin^2 x + \cos^2 x = 1$, substitute $\sin x = -2k$ and $\cos x = k$: $(-2k)^2 + k^2 = 1.$ $4k^2 + k^2 = 1.$ $5k^2 = 1.$ $k^2 = \frac{1}{5}.$ $k = \frac{1}{\sqrt{5}} \text{ or } k = -\frac{1}{\sqrt{5}}.$

Since $\cos x$ is positive in the 4th quadrant, we take $k = \frac{1}{\sqrt{5}}$. Thus: $\cos x = \frac{1}{\sqrt{5}} \quad \text{and} \quad \sin x = -\frac{2}{\sqrt{5}}.$

Step 4: Use the Double-Angle Formula for $\sin 2x$

The formula for $\sin 2x$ is: $\sin 2x = 2 \sin x \cos x.$

Substitute $\sin x = -\frac{2}{\sqrt{5}}$ and $\cos x = \frac{1}{\sqrt{5}}$: $\sin 2x = 2 \left(-\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{5}}\right).$ $\sin 2x = 2 \cdot -\frac{2}{5} = -\frac{4}{5}.$

Conclusion

The value of $\sin 2x$ is $-\frac{4}{5}$.

Answer

The correct answer is: (a) $-\frac{4}{5}$.

Would you like a deeper explanation of any step, or do you have further questions?

Related Questions

1. What are the values of $\cos 2x$ and $\tan 2x$ given the same conditions?
2. How would the solution change if $x$ were in the 2nd quadrant?
3. How is the double-angle formula for cosine derived?
4. Why does the sign of trigonometric functions vary across quadrants?
5. How does $\cot x$ relate to $\tan x$ and what transformations are involved?

Tip

When solving trigonometric problems, always pay attention to the quadrant since it affects the signs of sine, cosine, and tangent.