Let's solve each part by evaluating the cosine of the given angles.

(a) $\cos\left(\frac{3\pi}{4}\right)$

The angle $\frac{3\pi}{4}$ is in the second quadrant. In the second quadrant, cosine values are negative. We know that: $\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)$ Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we get: $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

(b) $\cos\left(\frac{5\pi}{4}\right)$

The angle $\frac{5\pi}{4}$ is in the third quadrant. In the third quadrant, cosine values are also negative. We know that: $\cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)$ Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we get: $\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

(c) $\cos\left(\frac{7\pi}{4}\right)$

The angle $\frac{7\pi}{4}$ is in the fourth quadrant. In the fourth quadrant, cosine values are positive. We have: $\cos\left(\frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$ Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we get: $\cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Summary of Answers:

• (a) $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
• (b) $\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
• (c) $\cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Would you like further explanation on any of these steps?

Here are some related questions to deepen your understanding:

1. How would you find the sine values for these angles?
2. Can you explain why cosine values are negative in the second and third quadrants?
3. What is the general formula for finding cosine values in different quadrants?
4. How would the answers change if we were working with $\sin$ instead of $\cos$?
5. Why is $\cos(\pi + x) = -\cos(x)$?

Tip: Remember that the unit circle is a powerful tool for evaluating trigonometric functions at standard angles.