(a) Show that for any angle θ \theta θ m m m ∣ sin m θ ∣ ≤ m ∣ sin θ ∣ | \sin m\theta| \le m| \sin \theta| ∣ sin m θ ∣ ≤ m ∣ sin θ ∣
1. Base Case: m = 1 m = 1 m = 1 ∣ sin θ ∣ = 1 ⋅ ∣ sin θ ∣ | \sin \theta | = 1 \cdot | \sin \theta | ∣ sin θ ∣ = 1 ⋅ ∣ sin θ ∣
2. Inductive Step: m = k m = k m = k ∣ sin k θ ∣ ≤ k ∣ sin θ ∣ | \sin k\theta | \le k | \sin \theta | ∣ sin k θ ∣ ≤ k ∣ sin θ ∣ m = k + 1 m = k + 1 m = k + 1
3. Using the Angle Addition Formula: sin ( ( k + 1 ) θ ) = sin ( k θ + θ ) = sin k θ cos θ + cos k θ sin θ \sin((k+1)\theta) = \sin(k\theta + \theta) = \sin k\theta \cos \theta + \cos k\theta \sin \theta sin (( k + 1 ) θ ) = sin ( k θ + θ ) = sin k θ cos θ + cos k θ sin θ
4. Taking the Absolute Value: ∣ sin ( ( k + 1 ) θ ) ∣ = ∣ sin k θ cos θ + cos k θ sin θ ∣ | \sin((k+1)\theta) | = | \sin k\theta \cos \theta + \cos k\theta \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ = ∣ sin k θ cos θ + cos k θ sin θ ∣
5. Applying the Triangle Inequality: ∣ sin ( ( k + 1 ) θ ) ∣ ≤ ∣ sin k θ cos θ ∣ + ∣ cos k θ sin θ ∣ | \sin((k+1)\theta) | \le | \sin k\theta \cos \theta | + | \cos k\theta \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ ∣ sin k θ cos θ ∣ + ∣ cos k θ sin θ ∣
6. **Using the Bound ∣ cos x ∣ ≤ 1 | \cos x | \le 1 ∣ cos x ∣ ≤ 1 ∣ sin ( ( k + 1 ) θ ) ∣ ≤ ∣ sin k θ ∣ ∣ cos θ ∣ + ∣ cos k θ ∣ ∣ sin θ ∣ | \sin((k+1)\theta) | \le | \sin k\theta | | \cos \theta | + | \cos k\theta | | \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ ∣ sin k θ ∣∣ cos θ ∣ + ∣ cos k θ ∣∣ sin θ ∣ ∣ sin ( ( k + 1 ) θ ) ∣ ≤ ∣ sin k θ ∣ ∣ cos θ ∣ + ∣ sin θ ∣ | \sin((k+1)\theta) | \le | \sin k\theta | | \cos \theta | + | \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ ∣ sin k θ ∣∣ cos θ ∣ + ∣ sin θ ∣
7. Using the Inductive Hypothesis: ∣ sin k θ ∣ ≤ k ∣ sin θ ∣ | \sin k\theta | \le k | \sin \theta | ∣ sin k θ ∣ ≤ k ∣ sin θ ∣ ∣ sin ( ( k + 1 ) θ ) ∣ ≤ k ∣ sin θ ∣ ∣ cos θ ∣ + ∣ sin θ ∣ | \sin((k+1)\theta) | \le k | \sin \theta | | \cos \theta | + | \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ k ∣ sin θ ∣∣ cos θ ∣ + ∣ sin θ ∣
8. Simplifying: ∣ sin ( ( k + 1 ) θ ) ∣ ≤ k ∣ sin θ ∣ + ∣ sin θ ∣ | \sin((k+1)\theta) | \le k | \sin \theta | + | \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ k ∣ sin θ ∣ + ∣ sin θ ∣ ∣ sin ( ( k + 1 ) θ ) ∣ ≤ ( k + 1 ) ∣ sin θ ∣ | \sin((k+1)\theta) | \le (k + 1) | \sin \theta | ∣ sin (( k + 1 ) θ ) ∣ ≤ ( k + 1 ) ∣ sin θ ∣
Thus, by induction, the inequality holds for all natural numbers m m m
■ \blacksquare ■
(b) Show that for all angles θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2 m m m ∣ sin m θ 2 − sin m θ 1 ∣ ≤ m ∣ sin ( θ 2 − θ 1 ) ∣ | \sin m \theta_2 - \sin m \theta_1| \le m| \sin (\theta_2 - \theta_1)| ∣ sin m θ 2 − sin m θ 1 ∣ ≤ m ∣ sin ( θ 2 − θ 1 ) ∣
1. **Let m = 2 k m = 2k m = 2 k k k k ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣
2. Using the Sum-to-Product Identities: sin A − sin B = 2 cos ( A + B 2 ) sin ( A − B 2 ) \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) sin A − sin B = 2 cos ( 2 A + B ) sin ( 2 A − B ) A = 2 k θ 2 A = 2k \theta_2 A = 2 k θ 2 B = 2 k θ 1 B = 2k \theta_1 B = 2 k θ 1 ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ = ∣ 2 cos ( k ( θ 2 + θ 1 ) ) sin ( k ( θ 2 − θ 1 ) ) ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | = | 2 \cos \left( k (\theta_2 + \theta_1) \right) \sin \left( k (\theta_2 - \theta_1) \right) | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ = ∣2 cos ( k ( θ 2 + θ 1 ) ) sin ( k ( θ 2 − θ 1 ) ) ∣
3. Taking the Absolute Value: ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ = 2 ∣ cos ( k ( θ 2 + θ 1 ) ) ∣ ∣ sin ( k ( θ 2 − θ 1 ) ) ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | = 2 | \cos \left( k (\theta_2 + \theta_1) \right) | | \sin \left( k (\theta_2 - \theta_1) \right) | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ = 2∣ cos ( k ( θ 2 + θ 1 ) ) ∣∣ sin ( k ( θ 2 − θ 1 ) ) ∣
4. **Using the Bound ∣ cos x ∣ ≤ 1 | \cos x | \le 1 ∣ cos x ∣ ≤ 1 ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ 2 ∣ sin ( k ( θ 2 − θ 1 ) ) ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | \le 2 | \sin \left( k (\theta_2 - \theta_1) \right) | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ 2∣ sin ( k ( θ 2 − θ 1 ) ) ∣
5. Using the Result from Part (a): ∣ sin k ( θ 2 − θ 1 ) ∣ ≤ k ∣ sin ( θ 2 − θ 1 ) ∣ | \sin k (\theta_2 - \theta_1) | \le k | \sin (\theta_2 - \theta_1) | ∣ sin k ( θ 2 − θ 1 ) ∣ ≤ k ∣ sin ( θ 2 − θ 1 ) ∣ ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ 2 k ∣ sin ( θ 2 − θ 1 ) ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | \le 2k | \sin (\theta_2 - \theta_1) | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ 2 k ∣ sin ( θ 2 − θ 1 ) ∣ ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ m ∣ sin ( θ 2 − θ 1 ) ∣ | \sin 2k \theta_2 - \sin 2k \theta_1 | \le m | \sin (\theta_2 - \theta_1) | ∣ sin 2 k θ 2 − sin 2 k θ 1 ∣ ≤ m ∣ sin ( θ 2 − θ 1 ) ∣
Thus, the inequality holds for all even natural numbers m m m
■ \blacksquare ■
(c) Show that for every odd natural number m m m θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2
1. **Consider m = 3 m = 3 m = 3 ∣ sin 3 θ 2 − sin 3 θ 1 ∣ ≤ 3 ∣ sin ( θ 2 − θ 1 ) ∣ | \sin 3 \theta_2 - \sin 3 \theta_1 | \le 3 | \sin (\theta_2 - \theta_1) | ∣ sin 3 θ 2 − sin 3 θ 1 ∣ ≤ 3∣ sin ( θ 2 − θ 1 ) ∣
2. **Choose θ 1 = 0 \theta_1 = 0 θ 1 = 0 θ 2 = π 3 \theta_2 = \frac{\pi}{3} θ 2 = 3 π ∣ sin 3 ⋅ π 3 − sin 0 ∣ = ∣ sin π − sin 0 ∣ = ∣ 0 − 0 ∣ = 0 | \sin 3 \cdot \frac{\pi}{3} - \sin 0 | = | \sin \pi - \sin 0 | = | 0 - 0 | = 0 ∣ sin 3 ⋅ 3 π − sin 0∣ = ∣ sin π − sin 0∣ = ∣0 − 0∣ = 0
3. Evaluate the Right-Hand Side: 3 ∣ sin ( π 3 − 0 ) ∣ = 3 ∣ sin π 3 ∣ = 3 ⋅ 3 2 = 3 3 2 3 | \sin (\frac{\pi}{3} - 0) | = 3 | \sin \frac{\pi}{3} | = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} 3∣ sin ( 3 π − 0 ) ∣ = 3∣ sin 3 π ∣ = 3 ⋅ 2 3 = 2 3 3
4. **Clearly, 0 ≤ 3 3 2 0 \le \frac{3\sqrt{3}}{2} 0 ≤ 2 3 3
5. Consider another pair of angles:**θ 1 = 0 \theta_1 = 0 θ 1 = 0 θ 2 = π 6 \theta_2 = \frac{\pi}{6} θ 2 = 6 π ∣ sin 3 ⋅ π 6 − sin 0 ∣ = ∣ sin π 2 − sin 0 ∣ = ∣ 1 − 0 ∣ = 1 | \sin 3 \cdot \frac{\pi}{6} - \sin 0 | = | \sin \frac{\pi}{2} - \sin 0 | = | 1 - 0 | = 1 ∣ sin 3 ⋅ 6 π − sin 0∣ = ∣ sin 2 π − sin 0∣ = ∣1 − 0∣ = 1
6. Evaluate the Right-Hand Side: 3 ∣ sin ( π 6 − 0 ) ∣ = 3 ∣ sin π 6 ∣ = 3 ⋅ 1 2 = 3 2 3 | \sin (\frac{\pi}{6} - 0) | = 3 | \sin \frac{\pi}{6} | = 3 \cdot \frac{1}{2} = \frac{3}{2} 3∣ sin ( 6 π − 0 ) ∣ = 3∣ sin 6 π ∣ = 3 ⋅ 2 1 = 2 3
7. **Clearly, 1 ≤ 3 2 1 \le \frac{3}{2} 1 ≤ 2 3
8. To show the inequality is invalid, consider the following:**θ 1 = 0 \theta_1 = 0 θ 1 = 0 θ 2 = π m \theta_2 = \frac{\pi}{m} θ 2 = m π ∣ sin m ⋅ π m − sin 0 ∣ = ∣ sin π − sin 0 ∣ = ∣ 0 − 0 ∣ = 0 | \sin m \cdot \frac{\pi}{m} - \sin 0 | = | \sin \pi - \sin 0 | = | 0 - 0 | = 0 ∣ sin m ⋅ m π − sin 0∣ = ∣ sin π − sin 0∣ = ∣0 − 0∣ = 0
9. Evaluate the Right-Hand Side: m ∣ sin ( π m − 0 ) ∣ = m ∣ sin π m ∣ m | \sin (\frac{\pi}{m} - 0) | = m | \sin \frac{\pi}{m} | m ∣ sin ( m π − 0 ) ∣ = m ∣ sin m π ∣
10. **For small θ \theta θ sin θ ≈ θ \sin \theta \approx \theta sin θ ≈ θ m ∣ sin π m ∣ ≈ m ⋅ π m = π m | \sin \frac{\pi}{m} | \approx m \cdot \frac{\pi}{m} = \pi m ∣ sin m π ∣ ≈ m ⋅ m π = π
11. **Clearly, 0 ≤ π 0 \le \pi 0 ≤ π
12. To show the inequality is invalid, consider the following:**θ 1 = 0 \theta_1 = 0 θ 1 = 0 θ 2 = π 2 m \theta_2 = \frac{\pi}{2m} θ 2 = 2 m π ∣ sin m ⋅ π 2 m − sin 0 ∣ = ∣ sin π 2 − sin 0 ∣ = ∣ 1 − 0 ∣ = 1 | \sin m \cdot \frac{\pi}{2m} - \sin 0 | = | \sin \frac{\pi}{2} - \sin 0 | = | 1 - 0 | = 1 ∣ sin m ⋅ 2 m π − sin 0∣ = ∣ sin 2 π − sin 0∣ = ∣1 − 0∣ = 1
13. Evaluate the Right-Hand Side: m ∣ sin ( π 2 m − 0 ) ∣ = m ∣ sin π 2 m ∣ m | \sin (\frac{\pi}{2m} - 0) | = m | \sin \frac{\pi}{2m} | m ∣ sin ( 2 m π − 0 ) ∣ = m ∣ sin 2 m π ∣
14. **For small θ \theta θ sin θ ≈ θ \sin \theta \approx \theta sin θ ≈ θ m ∣ sin π 2 m ∣ ≈ m ⋅ π 2 m = π 2 m | \sin \frac{\pi}{2m} | \approx m \cdot \frac{\pi}{2m} = \frac{\pi}{2} m ∣ sin 2 m π ∣ ≈ m ⋅ 2 m π = 2 π
15. **Clearly, 1 ≤ π 2 1 \le \frac{\pi}{2} 1 ≤ 2 π
Thus, for every odd natural number m m m θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2
■ \blacksquare ■