1. **Problem statement:** Calculate the following trigonometric problems: 6) a) Calculate $\sin 50^\circ + \sin 140^\circ + \sin 230^\circ + \sin 320^\circ$. b) Explain the result of a) without using a calculator by looking at the unit circle. c) Find possible angles $\alpha_2$, $\alpha_3$, and $\alpha_4$ such that $\sin 37^\circ + \sin \alpha_2 + \sin \alpha_3 + \sin \alpha_4 = 0$. 7) For each case, give the quadrant and two angles $\alpha$ where: a) $\sin \alpha > 0$ and $\cos \alpha > 0$ b) $\sin \alpha > 0$ and $\cos \alpha < 0$ c) $\sin \alpha < 0$ and $\cos \alpha > 0$ 8) Create a table of special sine and cosine values for angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, and their extensions by $90^\circ$, $180^\circ$, and $270^\circ$. --- 2. **Formulas and rules:** - Sine and cosine values on the unit circle correspond to the y- and x-coordinates respectively. - Sine and cosine are periodic with period $360^\circ$. - Quadrants are defined as: - Quadrant I: $0^\circ < \alpha < 90^\circ$ (sin > 0, cos > 0) - Quadrant II: $90^\circ < \alpha < 180^\circ$ (sin > 0, cos < 0) - Quadrant III: $180^\circ < \alpha < 270^\circ$ (sin < 0, cos < 0) - Quadrant IV: $270^\circ < \alpha < 360^\circ$ (sin < 0, cos > 0) --- 3. **Step 6a calculation:** Calculate each sine value: $\sin 50^\circ \approx 0.7660$ $\sin 140^\circ = \sin (180^\circ - 40^\circ) = \sin 40^\circ \approx 0.6428$ $\sin 230^\circ = \sin (180^\circ + 50^\circ) = -\sin 50^\circ \approx -0.7660$ $\sin 320^\circ = \sin (360^\circ - 40^\circ) = -\sin 40^\circ \approx -0.6428$ Sum: $$0.7660 + 0.6428 - 0.7660 - 0.6428 = 0$$ --- 4. **Step 6b explanation:** The positive sine values at $50^\circ$ and $140^\circ$ are canceled exactly by the negative sine values at $230^\circ$ and $320^\circ$ because these angles are symmetric about the $180^\circ$ and $360^\circ$ axes on the unit circle. --- 5. **Step 6c find angles:** Given $\sin 37^\circ + \sin \alpha_2 + \sin \alpha_3 + \sin \alpha_4 = 0$, choose angles symmetric to $37^\circ$ such that their sines sum to zero. Possible choices: $\alpha_2 = 143^\circ$ (since $\sin 143^\circ = \sin (180^\circ - 37^\circ) = \sin 37^\circ$) $\alpha_3 = 217^\circ$ (since $\sin 217^\circ = -\sin 37^\circ$) $\alpha_4 = 323^\circ$ (since $\sin 323^\circ = -\sin 37^\circ$) Sum: $$\sin 37^\circ + \sin 143^\circ + \sin 217^\circ + \sin 323^\circ = \sin 37^\circ + \sin 37^\circ - \sin 37^\circ - \sin 37^\circ = 0$$ --- 6. **Step 7 quadrants and angles:** a) $\sin \alpha > 0$ and $\cos \alpha > 0$ means Quadrant I. Two example angles: $30^\circ$, $60^\circ$ b) $\sin \alpha > 0$ and $\cos \alpha < 0$ means Quadrant II. Two example angles: $120^\circ$, $150^\circ$ c) $\sin \alpha < 0$ and $\cos \alpha > 0$ means Quadrant IV. Two example angles: $300^\circ$, $330^\circ$ --- 7. **Step 8 special values table:** | $\alpha$ | 0° | 30° | 45° | 60° | 90° | 120° | 135° | 150° | 180° | 210° | 225° | 240° | 270° | 300° | 315° | 330° | 360° | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | $\sin \alpha$ | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 | $-\frac{1}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{3}}{2}$ | $-1$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{1}{2}$ | 0 | | $\cos \alpha$ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 | $-\frac{1}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{3}}{2}$ | $-1$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{1}{2}$ | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 | --- **Final answers:** 6a) $0$ 6b) The sum is zero because the positive and negative sine values cancel out symmetrically on the unit circle. 6c) $\alpha_2 = 143^\circ$, $\alpha_3 = 217^\circ$, $\alpha_4 = 323^\circ$ 7a) Quadrant I, angles $30^\circ$, $60^\circ$ 7b) Quadrant II, angles $120^\circ$, $150^\circ$ 7c) Quadrant IV, angles $300^\circ$, $330^\circ$ 8) See table above for sine and cosine special values from $0^\circ$ to $360^\circ$.