# Question #49b10

Jan 20, 2018

$2 \sqrt{3} \cos \left(\theta + \frac{\pi}{3}\right)$

#### Explanation:

The general form is $a \cdot \cos \left(\theta\right) - b \cdot \sin \left(\theta\right) = R \cdot \cos \left(\theta + \alpha\right)$, where $R = \sqrt{{a}^{2} + {b}^{2}}$ and $\alpha = {\tan}^{-} 1 \left(\frac{b}{a}\right)$.

First calculate $R = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(\sqrt{3}\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{12} = 2 \sqrt{3}$.

Next calculate $\alpha = {\tan}^{-} 1 \left(\frac{b}{a}\right) = {\tan}^{-} 1 \left(\frac{3}{\sqrt{3}}\right) = {\tan}^{- 1} \left(\sqrt{3}\right) = \frac{\pi}{3}$.

So $\sqrt{3} \cos \left(\theta\right) - 3 \sin \left(\theta\right) = 2 \sqrt{3} \cos \left(\theta + \frac{\pi}{3}\right)$