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: $$R = \sqrt{a^2 + b^2}$$ $$\tan \alpha = \frac{b}{a}$$ where $R > 0$ and $\alpha$ is the phase shift. 3. Here, $a = 5$ and $b = -3$ (since the expression is $5 \sin x - 3 \cos x$). 4. Calculate $R$: $$R = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$$ 5. Calculate $\alpha$: $$\tan \alpha = \frac{-3}{5} = -0.6$$ Since $R > 0$ and $0 < \alpha < \frac{\pi}{2}$, and tangent is negative, we use: $$\alpha = \arctan\left(\frac{3}{5}\right) = 0.5404 \text{ radians (approx)}$$ 6. Therefore, the expression is: $$5 \sin x - 3 \cos x = \sqrt{34} \sin(x - 0.54)$$ --- 1. **Express $2 \cos(x - 60^\circ) + \cos x$ in the form $R \cos(x - \alpha)$** where $R > 0$ and $0^\circ < \alpha < 90^\circ$. 2. Expand $\cos(x - 60^\circ)$ using the cosine difference formula: $$\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ = \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x$$ 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 $R \cos(x - \alpha)$: 5. Use the formula: $$R = \sqrt{a^2 + b^2}$$ $$\tan \alpha = \frac{b}{a}$$ where $a = 2$ and $b = \sqrt{3}$. 6. Calculate $R$: $$R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}$$ 7. Calculate $\alpha$: $$\tan \alpha = \frac{\sqrt{3}}{2} \approx 0.8660$$ $$\alpha = \arctan(0.8660) = 0.715 \text{ radians} = 40.89^\circ$$ 8. 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: $$R = \sqrt{a^2 + b^2}$$ $$\tan \alpha = \frac{b}{a}$$ where $a = \sqrt{2}$ and $b = -\sqrt{5}$ (since the expression is $\sqrt{2} \cos x - \sqrt{5} \sin x$). 3. Calculate $R$: $$R = \sqrt{(\sqrt{2})^2 + (-\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}$$ 4. Calculate $\alpha$: $$\tan \alpha = \frac{-\sqrt{5}}{\sqrt{2}} = -\sqrt{\frac{5}{2}}$$ Since $\alpha$ is positive and between $0^\circ$ and $90^\circ$, and tangent is negative, we take: $$\alpha = \arctan\left(\sqrt{\frac{5}{2}}\right) = 1.150 \text{ radians (approx)}$$ 5. Final expression: $$\sqrt{2} \cos x - \sqrt{5} \sin x = \sqrt{7} \cos(x + 1.150)$$ --- 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: $$\sqrt{7} \cos(2\theta + 1.150) = 1$$ 3. Divide both sides by $\sqrt{7}$: $$\cos(2\theta + 1.150) = \frac{1}{\sqrt{7}}$$ 4. Find the principal value: $$2\theta + 1.150 = \arccos\left(\frac{1}{\sqrt{7}}\right) = 1.190 \text{ radians}$$ 5. General solutions for cosine: $$2\theta + 1.150 = 1.190 + 2k\pi \quad \text{or} \quad 2\theta + 1.150 = -1.190 + 2k\pi$$ 6. Solve for $\theta$: $$\theta = \frac{1.190 - 1.150 + 2k\pi}{2} = \frac{0.040 + 2k\pi}{2}$$ $$\theta = \frac{-1.190 - 1.150 + 2k\pi}{2} = \frac{-2.340 + 2k\pi}{2}$$ 7. Find values of $\theta$ in degrees between $0^\circ$ and $180^\circ$ (convert radians to degrees): - For $k=0$: $$\theta_1 = \frac{0.040}{2} = 0.020 \text{ radians} = 1.15^\circ$$ $$\theta_2 = \frac{-2.340}{2} = -1.170 \text{ radians} = -67.05^\circ \text{ (discard negative)}$$ - For $k=1$: $$\theta_3 = \frac{0.040 + 2\pi}{2} = \frac{0.040 + 6.283}{2} = 3.161 \text{ radians} = 181.1^\circ \text{ (discard > 180)}$$ $$\theta_4 = \frac{-2.340 + 2\pi}{2} = \frac{-2.340 + 6.283}{2} = 1.972 \text{ radians} = 112.99^\circ$$ 8. Valid solutions: $$\theta = 1.15^\circ, 112.99^\circ$$