To solve this problem, we need to determine the values of $\sin \theta$ and $\cos \theta$ based on the given point on the unit circle, $\left( \frac{6}{13}, y \right)$, in the first quadrant.

Step 1: Recall the equation of the unit circle

The equation of the unit circle is: $x^2 + y^2 = 1$ Given $x = \frac{6}{13}$, substitute $x^2$ into the equation: $\left( \frac{6}{13} \right)^2 + y^2 = 1$ $\frac{36}{169} + y^2 = 1$

Step 2: Solve for $y^2$

Rearrange to isolate $y^2$: $y^2 = 1 - \frac{36}{169}$ $y^2 = \frac{169}{169} - \frac{36}{169}$ $y^2 = \frac{133}{169}$

Step 3: Solve for $y$

Since $y$ is positive in the first quadrant: $y = \sqrt{\frac{133}{169}} = \frac{\sqrt{133}}{13}$

Step 4: Values of $\sin \theta$ and $\cos \theta$

From the unit circle: $\cos \theta = x = \frac{6}{13}, \quad \sin \theta = y = \frac{\sqrt{133}}{13}$

Final Answer:

$\cos \theta = \frac{6}{13}, \quad \sin \theta = \frac{\sqrt{133}}{13}$

Let me know if you'd like a detailed breakdown or further clarification!

Relative questions to deepen understanding:

1. How do the coordinates of a point on the unit circle relate to $\sin \theta$ and $\cos \theta$?
2. Why do we choose the positive square root for $y$ in the first quadrant?
3. What is the relationship between the Pythagorean theorem and the unit circle equation?
4. How would $\sin \theta$ and $\cos \theta$ differ if the point were in the second quadrant?
5. Can you verify $\cos^2 \theta + \sin^2 \theta = 1$ with these values?

Tip: Always check the quadrant when solving trigonometric problems to determine the sign of sine and cosine values!