Write $\cos\theta-\sqrt{3}\sin\theta$ in the form $r\sin(\theta-\alpha)$

Question

With this question I got

$\cos\alpha=-\sqrt{3}$ and $-r\sin\alpha=1$

thus $r\sin\alpha=-1$. Both of these are negative, so my solution should be in third quadrant. In the answer however, it is in first quadrant and I don't understand why. I got $\alpha= 210^\circ$ but the answer is $\alpha=30^\circ$.

Answer 1

$$\cos(\theta)-\sqrt 3 \sin(\theta) = 2\times (\frac{1}{2} \cos (\theta) - \frac{\sqrt 3}{2} \sin(\theta))$$

$$=2\times(\sin(\pi/6)\cos(\theta) - \cos(\pi/6)\sin(\theta))$$

$$=2\times(\sin(\pi/6-\theta))$$

$$=-2\times \sin(\theta-\pi/6)$$

Answer 2

Hint: $(\cos\theta -\sqrt3 \sin\theta)/2=\sin\left(\frac{\pi}{6}\right)\cos\theta-\cos\left(\frac{\pi}{6}\right)\sin\theta$. The reason for this is because they require the answer to be writen in the form of $\sin(\theta -\alpha)$, thus you need to rewrite the coefficient of $\sin\theta$ as $\cos\alpha$ to fit into the identity $\sin(a-b)=\sin a\cos b-\cos a \sin b$