We are given that $\sin(x) = -0.97$ and that $x$ is between $-\frac{\pi}{2}$ and $-\pi$. We are tasked with finding $\tan(x)$.

Step 1: Use the identity $\sin^2(x) + \cos^2(x) = 1$

We can find $\cos(x)$ using the Pythagorean identity:

$\sin^2(x) + \cos^2(x) = 1$

Substituting $\sin(x) = -0.97$:

$(-0.97)^2 + \cos^2(x) = 1$ $0.9409 + \cos^2(x) = 1$ $\cos^2(x) = 1 - 0.9409 = 0.0591$ $\cos(x) = \pm \sqrt{0.0591} \approx \pm 0.2431$

Step 2: Determine the sign of $\cos(x)$

Since $x$ is between $-\frac{\pi}{2}$ and $-\pi$, we are in the third quadrant, where both sine and cosine are negative. Therefore, $\cos(x) = -0.2431$.

Step 3: Find $\tan(x)$

The tangent function is given by:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

Substitute the values of $\sin(x)$ and $\cos(x)$:

$\tan(x) = \frac{-0.97}{-0.2431} \approx 3.99$

Thus, $\tan(x) \approx 3.99$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How can we find cosine directly from sine using other methods?
2. What is the behavior of the tangent function in different quadrants?
3. How do we determine the sign of trigonometric functions in each quadrant?
4. How can the unit circle help in solving this type of problem?
5. How does rounding affect the precision of trigonometric calculations?

Tip: Always pay attention to the quadrant to determine the correct signs of trigonometric functions.