1. **State the problem:** We need to define the three primary trigonometric ratios in terms of Cartesian coordinates $x$, $y$, and radius $r$. 2. **Primary trig ratios:** - $\sin \theta = \frac{y}{r}$ - $\cos \theta = \frac{x}{r}$ - $\tan \theta = \frac{y}{x}$ Here, $r = \sqrt{x^2 + y^2}$ is the distance from the origin to the point $(x,y)$. 3. **Signs of trig ratios in quadrants (CAST rule):** - Quadrant I: All positive - Quadrant II: Sine positive - Quadrant III: Tangent positive - Quadrant IV: Cosine positive 4. **Example 3: Determine sign of ratios without calculator:** (a) $\tan(\pi/5)$: $\pi/5 \approx 36^\circ$ in Quadrant I, tangent positive. (b) $\cos(3\pi/4)$: $3\pi/4 = 135^\circ$ in Quadrant II, cosine negative. (c) $\sin(5\pi/3)$: $5\pi/3 = 300^\circ$ in Quadrant IV, sine negative. (d) $\sec(3\pi/4)$: $\sec \theta = \frac{1}{\cos \theta}$, $3\pi/4$ in Quadrant II where cosine negative, so secant negative. 5. **Example 4: Exact values using special triangles and CAST:** (a) $\sin(5\pi/4)$: $5\pi/4 = 225^\circ$ in Quadrant III, reference angle $\pi/4$, sine negative. $$\sin(5\pi/4) = -\frac{\sqrt{2}}{2}$$ (b) $\cos(3\pi/4)$: Quadrant II, reference $\pi/4$, cosine negative. $$\cos(3\pi/4) = -\frac{\sqrt{2}}{2}$$ (c) $\tan(5\pi/3)$: $5\pi/3 = 300^\circ$ in Quadrant IV, reference $\pi/3$, tangent negative. $$\tan(5\pi/3) = -\sqrt{3}$$ (d) $\sin(7\pi/6)$: $7\pi/6 = 210^\circ$ in Quadrant III, reference $\pi/6$, sine negative. $$\sin(7\pi/6) = -\frac{1}{2}$$ (e) $\sec(5\pi/4)$: $5\pi/4$ in Quadrant III, cosine negative, so secant negative. $$\sec(5\pi/4) = -\sqrt{2}$$ 6. **Example 5: Solve $\cos \theta = -\frac{1}{\sqrt{2}}$ for $0 \leq \theta \leq 2\pi$:** - Reference angle $\theta_r = \frac{\pi}{4}$ - Cosine negative in Quadrants II and III - Solutions: $$\theta = \frac{3\pi}{4}, \frac{5\pi}{4}$$ 7. **Polar coordinates conversion:** For point $P(x,y)$, $$r = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\left(\frac{y}{x}\right)$$ 8. **Sign determination examples:** - $\tan(7\pi/4)$: Quadrant IV, tangent negative. - $\cos(5\pi/4)$: Quadrant III, cosine negative. - $\csc(2\pi/3)$: Quadrant II, sine positive, so cosecant positive. - $\sec(-\pi/6)$: $-\pi/6$ coterminal with $11\pi/6$ in Quadrant IV, cosine positive, secant positive. - $\sin(5\pi/3)$: Quadrant IV, sine negative. - $\sec(3\pi/2)$: $3\pi/2$ on negative y-axis, cosine zero, secant undefined. 9. **Exact values examples:** - $\cos(2\pi/3) = -\frac{1}{2}$ - $\sin(5\pi/4) = -\frac{\sqrt{2}}{2}$ - $\cot(3\pi/4) = -1$ - $\cos(3\pi/2) = 0$ - $\cos(-\pi/6) = \frac{\sqrt{3}}{2}$ - $\sin(\pi/2) = 1$ - $\sec(7\pi/4) = \sqrt{2}$ - $\tan(5\pi/3) = -\sqrt{3}$ - $\sin(7\pi/6) = -\frac{1}{2}$ - $\csc(-5\pi/5) = \csc(-\pi) = \text{undefined}$ (since $\sin(-\pi) = 0$) - $\cos(2\pi) = 1$ - $\tan(\pi/3) = \sqrt{3}$ 10. **Find all $\theta$ such that $0 \leq \theta \leq 2\pi$ for given equations:** (a) $\sin \theta = \frac{1}{2}$ - Reference angle $\frac{\pi}{6}$ - Solutions: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ (b) $\tan \theta = -1$ - Reference angle $\frac{\pi}{4}$ - Tangent negative in Quadrants II and IV - Solutions: $\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$ (c) $\sec \theta = \frac{2}{\sqrt{3}}$ - $\sec \theta = \frac{1}{\cos \theta}$, so $\cos \theta = \frac{\sqrt{3}}{2}$ - Reference angle $\frac{\pi}{6}$ - Cosine positive in Quadrants I and IV - Solutions: $\theta = \frac{\pi}{6}, \frac{11\pi}{6}$ (d) $\cos \theta = 0$ - Solutions: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ (e) $\cot \theta = \sqrt{3}$ - $\cot \theta = \frac{1}{\tan \theta}$, so $\tan \theta = \frac{1}{\sqrt{3}}$ - Reference angle $\frac{\pi}{6}$ - Tangent positive in Quadrants I and III - Solutions: $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$ (f) $\cos \theta = -\frac{1}{\sqrt{2}}$ - Reference angle $\frac{\pi}{4}$ - Cosine negative in Quadrants II and III - Solutions: $\theta = \frac{3\pi}{4}, \frac{5\pi}{4}$ (g) $\sin \theta = 0$ - Solutions: $\theta = 0, \pi, 2\pi$ (h) $\tan \theta = -\frac{1}{\sqrt{3}}$ - Reference angle $\frac{\pi}{6}$ - Tangent negative in Quadrants II and IV - Solutions: $\theta = \frac{5\pi}{6}, \frac{11\pi}{6}$ (i) $\csc \theta = -1$ - $\csc \theta = \frac{1}{\sin \theta}$, so $\sin \theta = -1$ - Solution: $\theta = \frac{3\pi}{2}$ 11. **Sum and difference of trig functions:** (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) $\cos 2\pi + \cos 0 = 1 + 1 = 2$ 12. **Find unknown side $x$ in triangle with angle $5\pi/6$ opposite $x$, sides 2 and $\sqrt{3}$:** Use Law of Sines: $$\frac{x}{\sin(5\pi/6)} = \frac{2}{\sin(\text{angle opposite 2})}$$ Since $\sin(5\pi/6) = \frac{1}{2}$, and angle opposite 2 is $\pi/3$ (since $\sqrt{3}$ side adjacent), $$x = 2 \times \frac{\sin(5\pi/6)}{\sin(\pi/3)} = 2 \times \frac{1/2}{\sqrt{3}/2} = 2 \times \frac{1/2}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ **Final answers:** - Primary trig ratios: $\sin \theta = \frac{y}{r}$, $\cos \theta = \frac{x}{r}$, $\tan \theta = \frac{y}{x}$ - Signs by quadrant: Quadrant I all positive, II sine positive, III tangent positive, IV cosine positive - Example 3 signs: (a) positive, (b) negative, (c) negative, (d) negative - Example 4 exact values: (a) $-\frac{\sqrt{2}}{2}$, (b) $-\frac{\sqrt{2}}{2}$, (c) $-\sqrt{3}$, (d) $-\frac{1}{2}$, (e) $-\sqrt{2}$ - Example 5 solutions for $\cos \theta = -\frac{1}{\sqrt{2}}$: $\frac{3\pi}{4}, \frac{5\pi}{4}$ - Unknown side $x = \frac{2\sqrt{3}}{3}$