To solve the given problem, we are provided with:

$\sin \theta = \frac{4}{5}, \quad \cos \theta < 0$

Steps:

1. Understand the Quadrant:
   Since $\sin \theta > 0$ and $\cos \theta < 0$, the angle $\theta$ is in the second quadrant (where sine is positive, and cosine is negative).

2. Find $\cos \theta$ using the Pythagorean identity:
   Recall the identity: $\sin^2 \theta + \cos^2 \theta = 1$ Substituting $\sin \theta = \frac{4}{5}$: $\left(\frac{4}{5}\right)^2 + \cos^2 \theta = 1$ $\frac{16}{25} + \cos^2 \theta = 1$ $\cos^2 \theta = 1 - \frac{16}{25} = \frac{9}{25}$ Therefore: $\cos \theta = \pm \frac{3}{5}$ Since $\cos \theta < 0$ (in the second quadrant): $\cos \theta = -\frac{3}{5}$

3. Check $\tan \theta$ (if needed):
   Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$: $\tan \theta = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}$

Final Answer:

• $\sin \theta = \frac{4}{5}$
• $\cos \theta = -\frac{3}{5}$
• $\tan \theta = -\frac{4}{3}$

Do you want details on any step, or have any questions?

Related Questions:

1. What are the values of $\sec \theta$, $\csc \theta$, and $\cot \theta$?
2. How does the quadrant affect the signs of trigonometric functions?
3. Can we solve this problem using a unit circle approach?
4. How is the Pythagorean identity derived?
5. How do we verify these results graphically?

Tip:

When determining signs of trigonometric functions, always consider the quadrant using the ASTC rule (All Students Take Calculus).