Trigonometry reference angle of radian

Question

Given that $-2\pi≤\theta≤0$ and $\theta$ has a reference angle of $\cfrac{\pi}{6}$ , find $\theta$ if it is in the

a) 1st quadrant

b) 2nd quadrant

c) 3rd quadrant

d) 4th quadrant

I need help on this problem which i'm unfamiliar with negative in radian..

Answer 1

Note that $$\sin \alpha = -\sin (-\alpha) = \sin(\pi - \alpha) = -\sin(\alpha - \pi)$$ and $$\cos \alpha = \cos (-\alpha) = -\cos(\pi - \alpha) = -\cos(\alpha - \pi).$$

If you try $\alpha = \frac{\pi}{6}$ and simplify the resulting formulas, you will find four angles that all have the same reference angle, $\frac{\pi}{6}$. If you then determine the quadrant in which each angle lies, you should find you have one angle in each quadrant.

Now your only remaining problem should be that some of the angles are not in the range you want ($-2\pi \le \theta \le 0$). What happens if you add or subtract $2\pi$ from one of those angles?

Answer 2

Hint:
Use the identities listed above: $$\cos(\theta+\pi)=-\cos\theta. \\ \sin(\theta+\pi)=-\sin\theta$$