1. Problem: Find the trigonometric ratios for given angles using the unit circle. 2. Formula: On the unit circle, coordinates at angle $\theta$ are $(\cos \theta, \sin \theta)$. 3. Important: $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$. 4. Calculate each: - $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$ - $\cos 270^\circ = 0$ - $\tan 60^\circ = \sqrt{3}$ - $\tan 300^\circ = \tan(360^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$ - $\sin 225^\circ = \sin(180^\circ + 45^\circ) = -\sin 45^\circ = -\frac{\sqrt{2}}{2}$ - $\tan 0^\circ = 0$ - $\cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$ - $\sin 225^\circ$ (repeat) $= -\frac{\sqrt{2}}{2}$ - $\cos 180^\circ = -1$ - $\sin 240^\circ = \sin(180^\circ + 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ - $\tan 270^\circ$ undefined (division by zero) - $\sin 180^\circ = 0$ - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ - $\sin 300^\circ = \sin(360^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ - $\tan 45^\circ = 1$ - $\csc 135^\circ = \frac{1}{\sin 135^\circ} = \frac{1}{\sin 45^\circ} = \sqrt{2}$ - $\sec 330^\circ = \frac{1}{\cos 330^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ - $\cot 360^\circ = \cot 0^\circ = \text{undefined (division by zero)}$ - $\cot 90^\circ = \frac{\cos 90^\circ}{\sin 90^\circ} = 0$ - $\sec 120^\circ = \frac{1}{\cos 120^\circ} = \frac{1}{-\frac{1}{2}} = -2$ - $\csc 240^\circ = \frac{1}{\sin 240^\circ} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ - $\cot 150^\circ = \frac{\cos 150^\circ}{\sin 150^\circ} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}$ - $\csc 300^\circ = \frac{1}{\sin 300^\circ} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ - $\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2$ - $\sec 210^\circ = \frac{1}{\cos 210^\circ} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ - $\cot 330^\circ = \frac{\cos 330^\circ}{\sin 330^\circ} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$ - $\cot 60^\circ = \frac{\cos 60^\circ}{\sin 60^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ - $\sec 180^\circ = \frac{1}{\cos 180^\circ} = \frac{1}{-1} = -1$ - $\cot 30^\circ = \frac{\cos 30^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ - $\csc 270^\circ = \frac{1}{\sin 270^\circ} = \frac{1}{-1} = -1$ II. Word Problem 1. Arm B angle: $120^\circ + 120^\circ = 240^\circ$ (3-fold symmetry, $360^\circ/3=120^\circ$) Coordinates: $(\cos 240^\circ, \sin 240^\circ) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ III. Pythagorean Theorem 1) Triangle ABC: Given $AB=12$, $AC=15$, find $BC$. $$BC = \sqrt{AC^2 - AB^2} = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9$$ 2) Triangle PQR: Given $PQ=5$, $PR=?$, $QR=?$ (assuming right triangle with missing side $QR$) If $PR$ is hypotenuse and $PQ$ one leg, missing side $QR = \sqrt{PR^2 - PQ^2}$ but $PR$ unknown, insufficient data. 3) Triangle UVW: Given $UV=6$, $VW=9$, find $UW$. $$UW = \sqrt{VW^2 - UV^2} = \sqrt{9^2 - 6^2} = \sqrt{81 - 36} = \sqrt{45} = 3\sqrt{5}$$ IV. Special Right Triangles 1) 45°-45°-90° triangle: legs equal, hypotenuse $x$, leg $2\sqrt{2}$ $$x = 2\sqrt{2} \times \sqrt{2} = 2 \times 2 = 4$$ $$y = 2\sqrt{2}$$ 2) 30°-60°-90° triangle: sides ratio $1: \sqrt{3}: 2$ Given base $20$ (opposite 60°), so base = $x\sqrt{3} = 20 \Rightarrow x = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}$ Hypotenuse $y = 2x = \frac{40\sqrt{3}}{3}$ 3) 30°-60°-90° triangle: given vertical side $3\sqrt{12} = 3 \times 2\sqrt{3} = 6\sqrt{3}$ (opposite 60°) Hypotenuse $x = \frac{6\sqrt{3}}{\sqrt{3}} \times 2 = 12$ Base $y = \frac{x}{2} = 6$ 4) 30°-60°-90° triangle: given top side $11\sqrt{3}$ (opposite 60°) Hypotenuse $x = \frac{11\sqrt{3}}{\sqrt{3}} \times 2 = 22$ Right side $y = \frac{x}{2} = 11$ 5) 45°-45°-90° triangle: hypotenuse $10$ Legs $x = y = \frac{10}{\sqrt{2}} = 5\sqrt{2}$ 6) 45°-45°-90° triangle: leg $6\sqrt{2}$ Hypotenuse $x = 6\sqrt{2} \times \sqrt{2} = 12$ Base $y = 6\sqrt{2}$ V. Trigonometric Ratios for triangle with legs 3 (opposite), 4 (adjacent), hypotenuse $x$ $$x = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$$ 1. $\sin \theta = \frac{3}{5}$ 2. $\cos \theta = \frac{4}{5}$ 3. $\tan \theta = \frac{3}{4}$ 4. $\sec \theta = \frac{5}{4}$ 5. $\csc \theta = \frac{5}{3}$ 6. $\cot \theta = \frac{4}{3}$