Let's work through both examples from the image you've provided.

Example 1:

Given:
$\cos \theta = -\frac{3}{5}$ and $\theta$ lies in Quadrant III.

Find: $\sin \theta$

In Quadrant III, both sine and cosine are negative, and we can use the Pythagorean identity:

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

Substituting the value of $\cos \theta = -\frac{3}{5}$:

$\sin^2 \theta + \left(-\frac{3}{5}\right)^2 = 1$ $\sin^2 \theta + \frac{9}{25} = 1$ $\sin^2 \theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}$ $\sin \theta = -\frac{4}{5}$

We take the negative root because $\sin \theta$ is negative in Quadrant III.

Answer: $\sin \theta = -\frac{4}{5}$

Example 2:

a. Evaluate $\cos 540^\circ + \sin 270^\circ$

First, let's reduce these angles:

• $540^\circ$ is coterminal with $540^\circ - 360^\circ = 180^\circ$, so $\cos 540^\circ = \cos 180^\circ = -1$.
• $\sin 270^\circ = -1$.

Now, calculate: $\cos 540^\circ + \sin 270^\circ = -1 + (-1) = -2$

Answer: $-2$

b. Evaluate $\cot 90^\circ + \tan(-90^\circ)$

• $\cot 90^\circ$ is undefined because $\cot \theta = \frac{1}{\tan \theta}$, and $\tan 90^\circ$ is undefined.
• $\tan(-90^\circ)$ is also undefined because it corresponds to a vertical asymptote on the unit circle.

Since both functions are undefined, the expression is undefined.

Would you like further clarification on these steps? Here are five follow-up questions to explore these concepts further:

1. What is the process to find the sine and cosine of other angles in different quadrants?
2. How do coterminal angles work in simplifying trigonometric expressions?
3. Why are some trigonometric functions undefined at certain angles?
4. Can you derive other trigonometric functions (like tangent or secant) from given values of sine or cosine?
5. How does the Pythagorean identity relate to solving for unknown trigonometric values?

Tip: When solving trigonometric problems, always consider the quadrant the angle lies in. This helps determine the correct sign of the sine and cosine values.