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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 $\alpha=30^\circ$.

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    $\begingroup$ 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. $\endgroup$
    – JMJ
    Apr 28, 2019 at 22:53

2 Answers 2

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

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

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

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

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    $\begingroup$ 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. $\endgroup$
    – N. F. Taussig
    Apr 28, 2019 at 22:58
  • $\begingroup$ If you want $r>0$ you need to add (or subtract) $\pi$, and the indeed, the angle is $210^\circ$ $\endgroup$
    – Andrei
    Apr 28, 2019 at 22:59
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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 $\sin(a-b)=\sin a\cos b-\cos a \sin b$

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