1. **Express 5 sin x - 3 cos x in the form $R \sin(x - \alpha)$** where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. 2. The formula to express $a \sin x + b \cos x$ in the form $R \sin(x - \alpha)$ is: $$a \sin x + b \cos x = R \sin(x - \alpha) = R(\sin x \cos \alpha - \cos x \sin \alpha)$$ where $R = \sqrt{a^2 + b^2}$ and $\tan \alpha = \frac{b}{a}$. 3. For $5 \sin x - 3 \cos x$, identify $a = 5$ and $b = -3$. Calculate $R$: $$R = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$$ Calculate $\alpha$: $$\tan \alpha = \frac{-3}{5} = -0.6$$ Since $R > 0$ and $0 < \alpha < \frac{\pi}{2}$, and tangent is negative, $\alpha$ is in the fourth quadrant, so we take $\alpha = \arctan\left(\frac{3}{5}\right)$ to keep $\alpha$ positive: $$\alpha = \arctan\left(\frac{3}{5}\right) \approx 0.5404 \text{ radians}$$ Rounded to 2 decimal places: $$\alpha \approx 0.54$$ 4. Final expression: $$5 \sin x - 3 \cos x = \sqrt{34} \sin(x - 0.54)$$ --- 1. **Show that $2 \cos(x - 60^\circ) + \cos x$ can be written as $R \cos(x - \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$.** 2. Expand $\cos(x - 60^\circ)$ using the formula: $$\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ$$ Substitute values: $$= \cos x \times \frac{1}{2} + \sin x \times \frac{\sqrt{3}}{2} = \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x$$ 3. Substitute into the expression: $$2 \cos(x - 60^\circ) + \cos x = 2 \left(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x\right) + \cos x = \cos x + \sqrt{3} \sin x + \cos x = 2 \cos x + \sqrt{3} \sin x$$ 4. Express $2 \cos x + \sqrt{3} \sin x$ in the form $R \cos(x - \alpha)$: $$R \cos(x - \alpha) = R(\cos x \cos \alpha + \sin x \sin \alpha)$$ where $$R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}$$ and $$\tan \alpha = \frac{\sqrt{3}}{2}$$ Calculate $\alpha$: $$\alpha = \arctan\left(\frac{\sqrt{3}}{2}\right) \approx 0.7137 \text{ radians} = 40.89^\circ$$ Rounded to 2 decimal places: $$\alpha \approx 40.89^\circ$$ 5. Final expression: $$2 \cos(x - 60^\circ) + \cos x = \sqrt{7} \cos(x - 40.89^\circ)$$ --- 1. **(a) Express $\sqrt{2} \cos x - \sqrt{5} \sin x$ in the form $R \cos(x + \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$.** 2. Use the formula: $$a \cos x + b \sin x = R \cos(x + \alpha) = R(\cos x \cos \alpha - \sin x \sin \alpha)$$ where $$R = \sqrt{a^2 + b^2}$$ and $$\tan \alpha = \frac{b}{a}$$ 3. Here, $a = \sqrt{2}$ and $b = -\sqrt{5}$. Calculate $R$: $$R = \sqrt{(\sqrt{2})^2 + (-\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}$$ Calculate $\alpha$: $$\tan \alpha = \frac{-\sqrt{5}}{\sqrt{2}} = -\sqrt{\frac{5}{2}}$$ Since $\alpha$ must be between $0^\circ$ and $90^\circ$, and tangent is negative, $\alpha$ is in the fourth quadrant, so take positive angle: $$\alpha = \arctan\left(\sqrt{\frac{5}{2}}\right) \approx 1.0304 \text{ radians} = 59.05^\circ$$ Rounded to 3 decimal places: $$\alpha \approx 1.030$$ 4. Final expression: $$\sqrt{2} \cos x - \sqrt{5} \sin x = \sqrt{7} \cos(x + 1.030)$$ --- 1. **(b) Solve $\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1$ for $0^\circ < \theta < 180^\circ$.** 2. Using the result from (a), rewrite the equation as: $$\sqrt{7} \cos(2\theta + 1.030) = 1$$ 3. Divide both sides by $\sqrt{7}$: $$\cos(2\theta + 1.030) = \frac{1}{\sqrt{7}}$$ 4. Find the principal value: $$2\theta + 1.030 = \arccos\left(\frac{1}{\sqrt{7}}\right) \approx 1.186$$ 5. General solutions for cosine: $$2\theta + 1.030 = 1.186 + 2k\pi \quad \text{or} \quad 2\theta + 1.030 = -1.186 + 2k\pi$$ for integers $k$. 6. Solve for $\theta$: $$\theta = \frac{1.186 - 1.030 + 2k\pi}{2} = \frac{0.156 + 2k\pi}{2}$$ $$\theta = \frac{-1.186 - 1.030 + 2k\pi}{2} = \frac{-2.216 + 2k\pi}{2}$$ 7. Convert radians to degrees ($1$ rad $\approx 57.2958^\circ$) and find values in $0^\circ < \theta < 180^\circ$: For $k=0$: $$\theta_1 = \frac{0.156}{2} = 0.078 \text{ rad} = 4.47^\circ$$ $$\theta_2 = \frac{-2.216}{2} = -1.108 \text{ rad} \text{ (discard negative)}$$ For $k=1$: $$\theta_3 = \frac{0.156 + 2\pi}{2} = \frac{0.156 + 6.283}{2} = 3.2195 \text{ rad} = 184.4^\circ \text{ (discard > 180)}$$ $$\theta_4 = \frac{-2.216 + 2\pi}{2} = \frac{-2.216 + 6.283}{2} = 2.0335 \text{ rad} = 116.5^\circ$$ 8. Valid solutions in $0^\circ < \theta < 180^\circ$ are approximately: $$\theta \approx 4.47^\circ, 116.5^\circ$$