1. **Problem statement:** (a) The population of a town is 5 million and it doubles every 15 years. Find the population after 20 years. 2. **Formula used:** The population growth formula given is $$P = P_0 \times 2^{\frac{t}{d}}$$ where: - $P_0$ is the initial population, - $t$ is the time elapsed, - $d$ is the doubling time. 3. **Applying the formula:** Given $P_0 = 5$ million, $t = 20$ years, and $d = 15$ years, $$P = 5 \times 2^{\frac{20}{15}} = 5 \times 2^{\frac{4}{3}}$$ 4. **Simplify the exponent:** $$2^{\frac{4}{3}} = 2^{1 + \frac{1}{3}} = 2^1 \times 2^{\frac{1}{3}} = 2 \times \sqrt[3]{2}$$ 5. **Calculate the population:** $$P = 5 \times 2 \times \sqrt[3]{2} = 10 \times \sqrt[3]{2}$$ million. 6. **Final answer:** The population after 20 years will be $$10 \sqrt[3]{2}$$ million. --- 1. **Problem statement:** (b) Find the exact value of $\sin 15^\circ$ using sum or difference identities. 2. **Formula used:** Use the difference identity for sine: $$\sin(a - b) = \sin a \cos b - \cos a \sin b$$ 3. **Apply the formula:** Let $a = 45^\circ$ and $b = 30^\circ$, so $$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$ 4. **Substitute known values:** $$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}$$ 5. **Calculate:** $$\sin 15^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$ 6. **Final answer:** $$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$$