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

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 $$\alpha=30^\circ$$.

• Hello and welcome to Math Stack Exchange! In the future, please formulate your question using MathJax (mathjax.org) as I have done for you here. – JMJ Apr 28 '19 at 22:53

$$\cos(\theta)-\sqrt 3 \sin(\theta) = 2\times (\frac{1}{2} \cos (\theta) - \frac{\sqrt 3}{2} \sin(\theta))$$
$$=2\times(\sin(\pi/6)\cos(\theta) - \cos(\pi/6)\sin(\theta))$$
$$=2\times(\sin(\pi/6-\theta))$$
$$=-2\times \sin(\theta-\pi/6)$$
• Formatting tip: To obtain $\sin\theta$, $\cos\theta$, $\tan\theta$, $\csc\theta$, $\sec\theta$, $\cot\theta$, type $\sin\theta$, $\cos\theta$, $\tan\theta$, $\csc\theta$, $\sec\theta$, $\cot\theta$, respectively. – N. F. Taussig Apr 28 '19 at 22:58
• If you want $r>0$ you need to add (or subtract) $\pi$, and the indeed, the angle is $210^\circ$ – Andrei Apr 28 '19 at 22:59
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 $$\sin(a-b)=\sin a\cos b-\cos a \sin b$$