1. **Problem Statement:** Evaluate the exact values of various trigonometric expressions and convert Cartesian coordinates to polar form, determine signs of trig ratios, and solve trig equations for angles between 0 and 2\pi. 2. **Formulas and Rules:** - Primary trig ratios: $\sin \theta = \frac{y}{r}$, $\cos \theta = \frac{x}{r}$, $\tan \theta = \frac{y}{x}$ where $r = \sqrt{x^2 + y^2}$. - Reciprocal ratios: $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{1}{\tan \theta}$. - Quadrant signs (CAST rule): Quadrant I all positive, II sin and csc positive, III tan and cot positive, IV cos and sec positive. - Polar coordinates: $r = \sqrt{x^2 + y^2}$, $\theta = \arctan(\frac{y}{x})$ adjusted by quadrant. 3. **Example 1:** a) $(\sin(\frac{\pi}{4}))(\cos(\frac{\pi}{4})) + (\sin(\frac{\pi}{6}))(\cos(\frac{\pi}{3}))$ - $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ - $\sin(\frac{\pi}{6}) = \frac{1}{2}$, $\cos(\frac{\pi}{3}) = \frac{1}{2}$ - Compute: $\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$ b) $\tan^2(\frac{\pi}{3}) - \sin^2(\frac{\pi}{4})$ - $\tan(\frac{\pi}{3}) = \sqrt{3}$, so $\tan^2(\frac{\pi}{3}) = 3$ - $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, so $\sin^2(\frac{\pi}{4}) = \frac{1}{2}$ - Compute: $3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2}$ c) $\csc^2(\frac{\pi}{3}) - \sec^2(\frac{\pi}{6})$ - $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, so $\csc(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}$, $\csc^2(\frac{\pi}{3}) = \frac{4}{3}$ - $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, so $\sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$, $\sec^2(\frac{\pi}{6}) = \frac{4}{3}$ - Compute: $\frac{4}{3} - \frac{4}{3} = 0$ 4. **Example 2:** Convert Cartesian to polar a) $P(1, \sqrt{3})$ - $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$ - $\theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$ - Polar: $(2, \frac{\pi}{3})$ b) $P(-\sqrt{3}, -1)$ - $r = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$ - $\theta = \arctan\left(\frac{-1}{-\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ but point is in quadrant III, so $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ - Polar: $(2, \frac{7\pi}{6})$ 5. **Example 3:** Sign of trig ratios a) $\tan(\frac{\pi}{5})$ in quadrant I, tan positive b) $\cos(\frac{3\pi}{4})$ quadrant II, cos negative c) $\sin(\frac{5\pi}{3})$ quadrant IV, sin negative d) $\sec(\frac{3\pi}{4})$ quadrant II, cos negative, so sec negative 6. **Example 4:** Exact values a) $\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$ (QIII) b) $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$ (QII) c) $\tan(\frac{5\pi}{3}) = -\sqrt{3}$ (QIV) d) $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$ (QIII) e) $\sec(\frac{5\pi}{4}) = -\sqrt{2}$ (QIII) 7. **Example 5:** Solve $0 \leq \theta \leq 2\pi$ a) $\cos \theta = -\frac{1}{\sqrt{2}}$ solutions: $\theta = \frac{3\pi}{4}, \frac{5\pi}{4}$ b) $\sin \theta = -\frac{\sqrt{3}}{2}$ solutions: $\theta = \frac{4\pi}{3}, \frac{5\pi}{3}$ c) $\tan \theta = \frac{3}{4}$ solutions: $\theta = \arctan(\frac{3}{4}), \pi + \arctan(\frac{3}{4})$ d) $\sec \theta = -\sqrt{2}$ means $\cos \theta = -\frac{1}{\sqrt{2}}$, same as (a) e) $\cos \theta = -0.62$ solutions in QII and QIII f) $\sin \vartheta = 0$ solutions: $\vartheta = 0, \pi, 2\pi$ 8. **Additional Problems:** 1. Convert Cartesian to Polar a) $P(3,3)$: $r=3\sqrt{2}$, $\theta=\frac{\pi}{4}$ b) $P(-2,4)$: $r=2\sqrt{5}$, $\theta=\arctan(-2)$ adjusted to QII c) $P(0,-5)$: $r=5$, $\theta=\frac{3\pi}{2}$ d) $P(1,-\sqrt{3})$: $r=2$, $\theta=\frac{5\pi}{3}$ 2. Sign determination a) $\tan(\frac{7\pi}{4})$ QIV tan negative b) $\cos(\frac{5\pi}{4})$ QIII cos negative c) $\csc(\frac{2\pi}{3})$ QII sin positive so csc positive d) $\sec(-\frac{\pi}{6})$ $=\sec(\frac{11\pi}{6})$ QIV cos positive so sec positive e) $\sin(\frac{5\pi}{3})$ QIV sin negative f) $\sec(\frac{3\pi}{2})$ undefined (cos=0) 3. Exact values a) $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ b) $\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$ c) $\cot(\frac{3\pi}{4}) = -1$ d) $\cos(\frac{3\pi}{2}) = 0$ e) $\cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ f) $\sin(\frac{\pi}{2}) = 1$ g) $\sec(\frac{7\pi}{4}) = \sqrt{2}$ h) $\tan(\frac{5\pi}{3}) = -\sqrt{3}$ i) $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$ j) $\csc(-\pi) = \csc(\pi) = \text{undefined}$ (sin=0) k) $\cos(2\pi) = 1$ l) $\tan(\frac{\pi}{3}) = \sqrt{3}$ 4. Solve for $\theta$ a) $\sin \theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ b) $\tan \theta = -1$: $\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$ c) $\sec \theta = \frac{2}{\sqrt{3}}$: $\cos \theta = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{6}, \frac{11\pi}{6}$ d) $\cos \theta = 0$: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ e) $\cot \theta = \sqrt{3}$: $\tan \theta = \frac{1}{\sqrt{3}}$, $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$ f) $\cos \theta = -\frac{1}{\sqrt{2}}$: $\theta = \frac{3\pi}{4}, \frac{5\pi}{4}$ g) $\sin \theta = 0$: $\theta = 0, \pi, 2\pi$ h) $\tan \theta = -\frac{1}{\sqrt{3}}$: $\theta = \frac{5\pi}{6}, \frac{11\pi}{6}$ i) $\csc \theta = -1$: $\sin \theta = -1$, $\theta = \frac{3\pi}{2}$ 5. Evaluate sums a) $\sin(\frac{\pi}{2}) + \sin(\frac{3\pi}{2}) = 1 + (-1) = 0$ b) $\cos(\frac{3\pi}{2}) - \cos(\pi) = 0 - (-1) = 1$ c) $\cos(2\pi) + \cos(0) = 1 + 1 = 2$ 6. Find unknown side in triangle with sides 2, $\sqrt{3}$ and angle $\frac{5\pi}{6}$ - Use Law of Cosines: $x^2 = 2^2 + (\sqrt{3})^2 - 2 \cdot 2 \cdot \sqrt{3} \cdot \cos(\frac{5\pi}{6})$ - $x^2 = 4 + 3 - 4\sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) = 7 + 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 7 + 4 \cdot \frac{3}{2} = 7 + 6 = 13$ - $x = \sqrt{13}$ **Final answers:** - Example 1: a) $\frac{3}{4}$, b) $\frac{5}{2}$, c) $0$ - Example 2: a) $(2, \frac{\pi}{3})$, b) $(2, \frac{7\pi}{6})$ - Example 3: a) positive, b) negative, c) negative, d) negative - Example 4: a) $-\frac{\sqrt{2}}{2}$, b) $-\frac{\sqrt{2}}{2}$, c) $-\sqrt{3}$, d) $-\frac{1}{2}$, e) $-\sqrt{2}$ - Example 5: see step 7 - Additional problems: see steps 8 and 9 - Unknown side: $\sqrt{13}$