1. **Express 5 sin x - 3 cos x in the form $R \sin(x - \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$.** The problem is to rewrite the expression $5 \sin x - 3 \cos x$ as $R \sin(x - \alpha)$. 2. **Formula and rules:** We use the identity: $$R \sin(x - \alpha) = R(\sin x \cos \alpha - \cos x \sin \alpha) = R \cos \alpha \sin x - R \sin \alpha \cos x$$ Matching coefficients: $$5 = R \cos \alpha$$ $$-3 = -R \sin \alpha \Rightarrow 3 = R \sin \alpha$$ 3. **Find $R$:** $$R = \sqrt{(R \cos \alpha)^2 + (R \sin \alpha)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$$ 4. **Find $\alpha$:** $$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{3}{5}$$ So, $$\alpha = \arctan\left(\frac{3}{5}\right) \approx 0.54 \text{ radians (2 decimal places)}$$ --- 5. **Express $2 \cos(x - 60^\circ) + \cos x$ in the form $R \cos(x - \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$.** 6. **Expand $\cos(x - 60^\circ)$:** $$\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ = \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x$$ 7. **Rewrite the expression:** $$2 \cos(x - 60^\circ) + \cos x = 2 \left(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x\right) + \cos x = \cos x + \sqrt{3} \sin x + \cos x = 2 \cos x + \sqrt{3} \sin x$$ 8. **Express in form $R \cos(x - \alpha)$:** Using identity: $$R \cos(x - \alpha) = R(\cos x \cos \alpha + \sin x \sin \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x$$ Matching coefficients: $$2 = R \cos \alpha$$ $$\sqrt{3} = R \sin \alpha$$ 9. **Find $R$:** $$R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}$$ 10. **Find $\alpha$:** $$\tan \alpha = \frac{\sqrt{3}}{2}$$ So, $$\alpha = \arctan\left(\frac{\sqrt{3}}{2}\right) \approx 0.71 \text{ radians} = 40.89^\circ \approx 40.89^\circ$$ --- 11. **(a) Express $\sqrt{2} \cos x - \sqrt{5} \sin x$ in the form $R \cos(x + \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$.** 12. **Use identity:** $$R \cos(x + \alpha) = R(\cos x \cos \alpha - \sin x \sin \alpha) = R \cos \alpha \cos x - R \sin \alpha \sin x$$ Matching coefficients: $$\sqrt{2} = R \cos \alpha$$ $$-\sqrt{5} = -R \sin \alpha \Rightarrow \sqrt{5} = R \sin \alpha$$ 13. **Find $R$:** $$R = \sqrt{(\sqrt{2})^2 + (\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}$$ 14. **Find $\alpha$:** $$\tan \alpha = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}}$$ So, $$\alpha = \arctan\left(\sqrt{\frac{5}{2}}\right) \approx 1.030 \text{ radians (3 decimal places)}$$ --- 15. **(b) Solve $\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1$ for $0^\circ < \theta < 180^\circ$.** Rewrite using part (a): $$\sqrt{7} \cos(2\theta + \alpha) = 1$$ Divide both sides by $\sqrt{7}$: $$\cos(2\theta + \alpha) = \frac{1}{\sqrt{7}}$$ 16. **Find general solutions:** $$2\theta + \alpha = \pm \arccos\left(\frac{1}{\sqrt{7}}\right) + 2k\pi, \quad k \in \mathbb{Z}$$ Calculate $\arccos\left(\frac{1}{\sqrt{7}}\right) \approx 1.19$ radians. 17. **Solve for $\theta$:** $$\theta = \frac{-\alpha \pm 1.19 + 2k\pi}{2}$$ Using $\alpha \approx 1.030$: For $k=0$: $$\theta_1 = \frac{-1.030 + 1.19}{2} = 0.08 \text{ radians} = 4.58^\circ$$ $$\theta_2 = \frac{-1.030 - 1.19}{2} = -1.11 \text{ radians (discard negative)}$$ For $k=1$: $$\theta_3 = \frac{-1.030 + 1.19 + 2\pi}{2} = \frac{0.16 + 6.283}{2} = 3.22 \text{ radians} = 184.5^\circ (\text{discard > 180}^\circ)$$ $$\theta_4 = \frac{-1.030 - 1.19 + 2\pi}{2} = \frac{-2.22 + 6.283}{2} = 2.03 \text{ radians} = 116.3^\circ$$ 18. **Valid solutions in $0^\circ < \theta < 180^\circ$ are:** $$\theta \approx 4.58^\circ, 116.3^\circ$$