Let's solve this step by step. The problem requires finding $\sin 2x$, $\cos 2x$, and $\tan 2x$ given that $\sin x = -\frac{4}{5}$ and $x$ terminates in quadrant IV.

Step 1: Understand the given information

• $\sin x = -\frac{4}{5}$: Using the Pythagorean identity, we know: $\sin^2 x + \cos^2 x = 1.$ Substituting $\sin x = -\frac{4}{5}$: $\left(-\frac{4}{5}\right)^2 + \cos^2 x = 1.$ Simplify: $\frac{16}{25} + \cos^2 x = 1.$ $\cos^2 x = 1 - \frac{16}{25} = \frac{9}{25}.$ $\cos x = \pm \frac{3}{5}.$ Since $x$ is in quadrant IV, $\cos x > 0$, so: $\cos x = \frac{3}{5}.$

Step 2: Use double-angle identities

1. $\sin 2x$: $\sin 2x = 2 \sin x \cos x.$ Substitute $\sin x = -\frac{4}{5}$ and $\cos x = \frac{3}{5}$: $\sin 2x = 2 \cdot \left(-\frac{4}{5}\right) \cdot \frac{3}{5}.$ $\sin 2x = 2 \cdot -\frac{12}{25} = -\frac{24}{25}.$

2. $\cos 2x$: $\cos 2x = \cos^2 x - \sin^2 x.$ Substitute $\cos^2 x = \frac{9}{25}$ and $\sin^2 x = \frac{16}{25}$: $\cos 2x = \frac{9}{25} - \frac{16}{25}.$ $\cos 2x = -\frac{7}{25}.$

3. $\tan 2x$: $\tan 2x = \frac{\sin 2x}{\cos 2x}.$ Substitute $\sin 2x = -\frac{24}{25}$ and $\cos 2x = -\frac{7}{25}$: $\tan 2x = \frac{-\frac{24}{25}}{-\frac{7}{25}}.$ Simplify: $\tan 2x = \frac{24}{7}.$

Final Answers:

• $\sin 2x = -\frac{24}{25}$,
• $\cos 2x = -\frac{7}{25}$,
• $\tan 2x = \frac{24}{7}$.

Would you like more explanation on any of the steps?

Related Questions:

1. How do the quadrant rules affect the signs of $\sin$, $\cos$, and $\tan$?
2. What are the double-angle identities, and how are they derived?
3. Can you calculate $\tan x$ from the given information?
4. How do you verify the solutions using Pythagoras' theorem?
5. What happens to the double-angle results in other quadrants?

Tip:

Remember, when working in specific quadrants, always check the sign of $\sin x$, $\cos x$, and $\tan x$ carefully based on the angle's location.