Write the following as a single trigonometric ratio: (1) \( \frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ} \)

To express \(\frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ}\) as a single trigonometric ratio, we will recognize this as a linear combination of cosine and sine. We can rewrite it in the form \(R \cos(\theta + \phi)\), where: - \( R = \sqrt{(a^2 + b^2)} \) - \( a = \frac{1}{2} \) - \( b = -\frac{\sqrt{3}}{2} \) Calculating \(R\): \[ R = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \] Next, we find the angle \(\phi\): \[ \cos \phi = \frac{a}{R} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \quad \Rightarrow \quad \phi = 60^{\circ} \] \[ \sin \phi = \frac{b}{R} = \frac{-\frac{\sqrt{3}}{2}}{1} = -\frac{\sqrt{3}}{2} \quad \Rightarrow \quad \phi = -60^{\circ} \] Now, consolidating our components, we express \( \frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ} \) as: \[ \cos(10^{\circ} + 60^{\circ}) \text{ or } \cos(10^{\circ} - 60^{\circ}) \text{ depending on the context, but commonly it's expressed as } \cos(70^{\circ}) \] Thus, the single trigonometric ratio is: \[ \cos(70^{\circ}) \]