1. **Show that** $\sin 3x \equiv 3 \sin x - 4 \sin^3 x$. Start from the triple angle identity for sine: $$\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x.$$ Using double angle formulas: $$\sin 2x = 2 \sin x \cos x, \quad \cos 2x = 1 - 2 \sin^2 x.$$ Substitute these: $$\sin 3x = (2 \sin x \cos x)(\cos x) + (1 - 2 \sin^2 x)(\sin x).$$ Simplify: $$= 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x.$$ Recall that $\cos^2 x = 1 - \sin^2 x$, so $$2 \sin x (1 - \sin^2 x) + \sin x - 2 \sin^3 x = 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x.$$ Combine like terms: $$= 3 \sin x - 4 \sin^3 x.$$ Therefore, $$\sin 3x = 3 \sin x - 4 \sin^3 x.$$ 2. **Solve for** $0 \leq \theta < \pi$: $$8 \sin^3 x - 6 \sin x + 1 = 0.$$ Rewrite: $$8 \sin^3 x - 6 \sin x + 1 = 0.$$ Divide both sides by 2: $$4 \sin^3 x - 3 \sin x + \frac{1}{2} = 0.$$ Recall from part (1): $$\sin 3x = 3 \sin x - 4 \sin^3 x \implies -\sin 3x = 4 \sin^3 x - 3 \sin x.$$ Rewrite the equation as: $$-\sin 3x + \frac{1}{2} = 0 \implies \sin 3x = \frac{1}{2}.$$ Solve for $3x$: For $\sin \theta = \frac{1}{2}$, $\theta = \frac{\pi}{6} + 2k\pi$ or $\frac{5\pi}{6} + 2k\pi$ for integers $k$. Thus, $$3x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 3x = \frac{5\pi}{6} + 2k\pi.$$ Divide both sides by 3: $$x = \frac{\pi}{18} + \frac{2k\pi}{3} \quad \text{or} \quad x = \frac{5\pi}{18} + \frac{2k\pi}{3}.$$ Find all $x$ in $[0, \pi)$ by choosing $k$ appropriately: For $x = \frac{\pi}{18} + \frac{2k\pi}{3}$: - $k=0$: $x=\frac{\pi}{18}$ - $k=1$: $x=\frac{\pi}{18} + \frac{2\pi}{3} = \frac{13\pi}{18}$ - $k=2$: $x=\frac{\pi}{18} + \frac{4\pi}{3} > \pi$ (exclude) For $x = \frac{5\pi}{18} + \frac{2k\pi}{3}$: - $k=0$: $x=\frac{5\pi}{18}$ - $k=1$: $x=\frac{5\pi}{18} + \frac{2\pi}{3} = \frac{17\pi}{18}$ - $k=2$: $x= \frac{5\pi}{18} + \frac{4\pi}{3} > \pi$ (exclude) **Solutions:** $$x = \frac{\pi}{18}, \frac{13\pi}{18}, \frac{5\pi}{18}, \frac{17\pi}{18}.$$ 3. **Express** $2 \sin x + 3 \cos x$ **in the form** $R \sin(x + \alpha)$ where $R > 0$, $0 < \alpha < 90^\circ$. Recall the identity: $$R \sin(x + \alpha) = R \sin x \cos \alpha + R \cos x \sin \alpha.$$ Matching coefficients from $2 \sin x + 3 \cos x$: $$R \cos \alpha = 2, \quad R \sin \alpha = 3.$$ Find $R$: $$R = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.$$ Find $\alpha$: $$\tan \alpha = \frac{3}{2}, \quad \alpha = \tan^{-1} \left(\frac{3}{2}\right).$$ Calculate $\alpha$ (in degrees or radians as needed): $$\alpha \approx 56.31^\circ.$$ Thus, $$2 \sin x + 3 \cos x = \sqrt{13} \sin \left(x + 56.31^\circ \right).$$ 4. Temperature model: $$\theta = 8 + 2 \sin(15t - 160) + 3 \cos(15t - 160).$$ Using part (3), express the sinusoidal terms as: $$2 \sin(15t - 160) + 3 \cos(15t - 160) = \sqrt{13} \sin \left(15t - 160 + \alpha\right),$$ where $\alpha = \tan^{-1}\left(\frac{3}{2}\right) \approx 56.31^\circ.$ 5. **Find the maximum temperature:** The sine function ranges from $-1$ to $1$. The maximum of $\sin(15t - 160 + \alpha)$ is 1. Hence, $$\theta_{max} = 8 + \sqrt{13} \times 1 = 8 + \sqrt{13} \approx 8 + 3.60555 = 11.60555.$$ Rounded, $$\theta_{max} \approx 11.61^\circ C.$$ 6. **Find the time of day when the maximum temperature occurs.** Maximum occurs when $$\sin(15t - 160 + \alpha) = 1,$$ which implies $$15t - 160 + \alpha = 90^\circ + 360^\circ k, \quad k \in \mathbb{Z}.$$ Use degrees (since angles given in degrees): $$15t = 90 + 160 - \alpha + 360k = 250 - 56.31 + 360k = 193.69 + 360k.$$ For smallest non-negative $t$, take $k=0$: $$t = \frac{193.69}{15} = 12.9127 \text{ hours}.$$ Convert fractional hours into minutes: $$0.9127 \times 60 = 54.76 \approx 55 \text{ minutes}.$$ Thus, $$t \approx 12 \text{h } 55 \text{min}.$$ **Final answers:** - (a) $\sin 3x = 3 \sin x - 4 \sin^3 x$ - (b) $x = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$ - (17a) $2 \sin x + 3 \cos x = \sqrt{13} \sin \left(x + 56.31^\circ \right)$ - (17b) Maximum temperature $\approx 11.61^\circ C$ - (17c) Maximum temperature occurs at approximately 12:55 PM