Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $

Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $

Can someone please explain to me how to go about doing this?

you can write $$\sqrt 3 \sin t - \cos t = 2\left(\frac{\sqrt 3}2 \sin t - \frac 12 \cos t\right) = 2\left(\sin t\sin \left(\frac{\pi}3 \right) - \cos\left(\frac {\pi} 3\right) \cos t\right)= -2\cos \left(t + \frac{\pi}3\right) = 2 \cos \left(t - \frac{2\pi}3\right) $$

The moment you see $a$ and $b$ as coefficients of $\cos \theta , \sin \theta$,multiply and divide by $\sqrt { a^2 + b^2} $ so that the latter can be absorbed into $\theta $ additively. Now since $ \sqrt{ 3 +1} =2, $

$$\sqrt 3 \sin t - \cos t = 2\left(\sin t \frac{\sqrt 3}2 - \cos t \;\frac 12\right) = 2 \sin ( t - \pi/6).$$