Given that $\theta_1$ is located in Quadrant II and $\sin(\theta_1) = \dfrac{9}{41}$, we can find additional trigonometric values for $\theta_1$. Here's the process:

Step 1: Understand the Sine Function

• $\sin(\theta_1) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{9}{41}$
• In Quadrant II, sine is positive, and cosine is negative.

Step 2: Use the Pythagorean Identity to Find $\cos(\theta_1)$

The Pythagorean identity is:

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

Substitute $\sin(\theta_1) = \dfrac{9}{41}$:

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

$\dfrac{81}{1681} + \cos^2(\theta_1) = 1$

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

Step 3: Solve for $\cos(\theta_1)$

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

Since $\theta_1$ is in Quadrant II, $\cos(\theta_1)$ is negative:

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

Step 4: Find $\tan(\theta_1)$

The tangent is defined as:

$\tan(\theta_1) = \dfrac{\sin(\theta_1)}{\cos(\theta_1)}$

Substitute the values:

$\tan(\theta_1) = \dfrac{9/41}{-40/41} = -\dfrac{9}{40}$

Summary of Trigonometric Values:

• $\sin(\theta_1) = \dfrac{9}{41}$
• $\cos(\theta_1) = -\dfrac{40}{41}$
• $\tan(\theta_1) = -\dfrac{9}{40}$

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. How do the trigonometric identities change across different quadrants?
2. What is the relationship between sine, cosine, and tangent in general?
3. Can you derive the secant, cosecant, and cotangent values from this information?
4. How would the values change if $\theta_1$ were in Quadrant III?
5. Why is the cosine value negative in Quadrant II?

Tip: In trigonometry, always be mindful of the signs of the functions based on the quadrant of the angle.