The image provides a problem related to the Pythagorean Identity in trigonometry, which states:

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

Given that $\cos \theta = \frac{9}{10}$, we are asked to find $\sin \theta$.

Step-by-Step Solution:

1. Start with the Pythagorean Identity: $\sin^2 \theta + \cos^2 \theta = 1$

2. Substitute the given value for $\cos \theta$: $\sin^2 \theta + \left(\frac{9}{10}\right)^2 = 1$ $\sin^2 \theta + \frac{81}{100} = 1$

3. Solve for $\sin^2 \theta$: $\sin^2 \theta = 1 - \frac{81}{100}$ $\sin^2 \theta = \frac{100}{100} - \frac{81}{100}$ $\sin^2 \theta = \frac{19}{100}$

4. Now, take the square root of both sides to find $\sin \theta$: $\sin \theta = \sqrt{\frac{19}{100}} = \frac{\sqrt{19}}{10}$

Thus, the simplified value of $\sin \theta$ is:

$\sin \theta = \frac{\sqrt{19}}{10}$

Would you like further details or have any questions?

Here are 5 questions related to this concept:

1. How do we handle negative values of $\sin \theta$ based on the quadrant?
2. Can you derive the Pythagorean Identity from the unit circle?
3. What happens to the Pythagorean Identity if $\cos \theta = 0$?
4. How would the solution change if $\cos \theta$ were negative?
5. Can the Pythagorean Identity be extended to non-right triangles?

Tip:

In trigonometric identities, the signs of $\sin \theta$ and $\cos \theta$ depend on the quadrant in which the angle $\theta$ lies.