Let's solve each part step by step.

Given: $\theta = \frac{3\pi}{4}$

1. (a) $\cos(-\theta)$

   Using the even property of the cosine function, $\cos(-\theta) = \cos(\theta)$. So: $\cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right)$ The value of $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$.

   Therefore, $\cos(-\theta) = -\frac{\sqrt{2}}{2}$.

2. (b) $\cos^2 \theta$

   $\cos^2 \theta$ means $(\cos \theta)^2$. $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ So, $\cos^2 \theta = \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}$

3. (c) $\cos 2\theta$

   Using the double-angle formula, $\cos 2\theta = 2\cos^2 \theta - 1$. We already found $\cos^2 \theta = \frac{1}{2}$, so: $\cos 2\theta = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0$

Final Answers:

• (a) $\cos(-\theta) = -\frac{\sqrt{2}}{2}$
• (b) $\cos^2 \theta = \frac{1}{2}$
• (c) $\cos 2\theta = 0$

Would you like a more detailed breakdown of each step?

Here are 5 related questions for practice:

1. What is $\sin(-\theta)$ if $\theta = \frac{3\pi}{4}$?
2. Calculate $\tan(\theta)$ for $\theta = \frac{3\pi}{4}$.
3. Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to verify $\cos^2 \theta$.
4. How does the value of $\cos(2\theta)$ change if $\theta = \frac{\pi}{3}$ instead?
5. What is the value of $\sec(\theta)$ if $\theta = \frac{3\pi}{4}$?

Tip: For angles like $\frac{3\pi}{4}$, remember they fall in the second quadrant, where cosine values are negative.