Given that $\theta_1$ is in Quadrant II and $\sin(\theta_1) = \frac{9}{41}$, we can determine $\cos(\theta_1)$ using the Pythagorean identity:

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

Step 1: Express $\cos^2(\theta_1)$ in terms of $\sin(\theta_1)$:

$\cos^2(\theta_1) = 1 - \sin^2(\theta_1)$

Step 2: Substitute the given value $\sin(\theta_1) = \frac{9}{41}$ into the equation:

$\cos^2(\theta_1) = 1 - \left(\frac{9}{41}\right)^2$

Step 3: Calculate $\left(\frac{9}{41}\right)^2$:

$\left(\frac{9}{41}\right)^2 = \frac{81}{1681}$

Step 4: Subtract from 1:

$\cos^2(\theta_1) = 1 - \frac{81}{1681} = \frac{1681}{1681} - \frac{81}{1681} = \frac{1600}{1681}$

Step 5: Take the square root to find $\cos(\theta_1)$:

$\cos(\theta_1) = \pm\frac{\sqrt{1600}}{41} = \pm\frac{40}{41}$

Step 6: Determine the correct sign for $\cos(\theta_1)$ in Quadrant II:

In Quadrant II, cosine is negative, so:

$\cos(\theta_1) = -\frac{40}{41}$

Final Answer

$\cos(\theta_1) = -\frac{40}{41}$

Would you like more details or have any questions?

---

Here are 5 related questions:

1. If $\theta$ is in Quadrant III and $\sin(\theta) = -\frac{3}{5}$, what is $\cos(\theta)$?
2. How does the Pythagorean identity relate to trigonometric functions?
3. Why is cosine negative in Quadrant II?
4. How do you determine the sign of trigonometric functions based on the quadrant?
5. What is the exact value of $\cos(\theta)$ if $\sin(\theta) = \frac{12}{13}$ and $\theta$ is in Quadrant I?

Tip: Remember that in each quadrant, different trigonometric functions are positive or negative depending on the angle's location.