Trig Values Ed1Fe6

1. **State the problem:** We are given $\theta = 20^\circ$ and need to find $\alpha$ such that $0^\circ < \alpha < 360^\circ$, $\alpha \neq \theta$, and: a. $\sin(\alpha) = \sin(\theta)$ b. $\cos(\alpha) = \cos(\theta)$ Then compute exact values for: c. $\sin(-3\pi/4)$ d. $\cos(-3\pi/4)$ --- 2. **Recall important formulas and rules:** - For sine, $\sin(\alpha) = \sin(\theta)$ implies $\alpha = \theta$ or $\alpha = 180^\circ - \theta$ (in degrees). - For cosine, $\cos(\alpha) = \cos(\theta)$ implies $\alpha = \theta$ or $\alpha = 360^\circ - \theta$. - Exact values for sine and cosine at special angles like $-3\pi/4$ can be found using unit circle properties. --- 3. **Solve part a:** Given $\sin(\alpha) = \sin(20^\circ)$ and $\alpha \neq 20^\circ$, the other solution in $0^\circ < \alpha < 360^\circ$ is: $$\alpha = 180^\circ - 20^\circ = 160^\circ$$ --- 4. **Solve part b:** Given $\cos(\alpha) = \cos(20^\circ)$ and $\alpha \neq 20^\circ$, the other solution in $0^\circ < \alpha < 360^\circ$ is: $$\alpha = 360^\circ - 20^\circ = 340^\circ$$ --- 5. **Compute part c:** Calculate $\sin(-3\pi/4)$. Recall that $\sin(-x) = -\sin(x)$, so: $$\sin(-3\pi/4) = -\sin(3\pi/4)$$ Since $\sin(3\pi/4) = \sin(\pi - \pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}$, we get: $$\sin(-3\pi/4) = -\frac{\sqrt{2}}{2}$$ --- 6. **Compute part d:** Calculate $\cos(-3\pi/4)$. Recall that $\cos(-x) = \cos(x)$, so: $$\cos(-3\pi/4) = \cos(3\pi/4)$$ Since $\cos(3\pi/4) = \cos(\pi - \pi/4) = -\cos(\pi/4) = -\frac{\sqrt{2}}{2}$, we get: $$\cos(-3\pi/4) = -\frac{\sqrt{2}}{2}$$ --- **Final answers:** a. $\alpha = 160^\circ$ b. $\alpha = 340^\circ$ c. $\sin(-3\pi/4) = -\frac{\sqrt{2}}{2}$ d. $\cos(-3\pi/4) = -\frac{\sqrt{2}}{2}$