1. **Problem a:** Find the amplitude, period, and phase shift for $y = 2 \cos(3x + \frac{\pi}{2})$ and sketch the graph for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$. 2. **Amplitude:** The amplitude of $y = a \cos(bx + c)$ is $|a|$. Here, $a = 2$, so amplitude = $2$. 3. **Period:** The period is given by $\frac{2\pi}{|b|}$. Here, $b = 3$, so period = $\frac{2\pi}{3}$. 4. **Phase shift:** The phase shift is $-\frac{c}{b}$. Here, $c = \frac{\pi}{2}$, so phase shift = $-\frac{\frac{\pi}{2}}{3} = -\frac{\pi}{6}$. This means the graph shifts $\frac{\pi}{6}$ units to the left. 5. **Summary for a:** Amplitude = $2$, Period = $\frac{2\pi}{3}$, Phase shift = $-\frac{\pi}{6}$. --- 6. **Problem b:** Find amplitude, period, and phase shift for $f(x) = -3 \sin(2x + \frac{\pi}{2})$ and sketch for $-\pi \leq x \leq \pi$. 7. **Amplitude:** $|a| = |-3| = 3$. 8. **Period:** $\frac{2\pi}{|b|} = \frac{2\pi}{2} = \pi$. 9. **Phase shift:** $-\frac{c}{b} = -\frac{\frac{\pi}{2}}{2} = -\frac{\pi}{4}$. Shift $\frac{\pi}{4}$ units left. 10. **Summary for b:** Amplitude = $3$, Period = $\pi$, Phase shift = $-\frac{\pi}{4}$. --- 11. **Problem c:** Given $\sin \theta = \frac{\sqrt{21}}{5}$ and $\frac{\pi}{2} \leq \theta \leq \pi$, find $\sin 2\theta$ and $\cos 2\theta$ without a calculator. 12. Since $\theta$ is in quadrant II, $\sin \theta > 0$ and $\cos \theta < 0$. 13. Use Pythagoras: $\sin^2 \theta + \cos^2 \theta = 1$. $$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \left(\frac{\sqrt{21}}{5}\right)^2} = -\sqrt{1 - \frac{21}{25}} = -\sqrt{\frac{4}{25}} = -\frac{2}{5}$$ 14. Use double angle formulas: $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{\sqrt{21}}{5} \times \left(-\frac{2}{5}\right) = -\frac{4\sqrt{21}}{25}$$ $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(-\frac{2}{5}\right)^2 - \left(\frac{\sqrt{21}}{5}\right)^2 = \frac{4}{25} - \frac{21}{25} = -\frac{17}{25}$$ 15. **Final answers:** $\sin 2\theta = -\frac{4\sqrt{21}}{25}$, $\cos 2\theta = -\frac{17}{25}$