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..

You surely have a typo in your first sentence, do you mean $-2\pi\leqslant\theta\leqslant0$ or something like that? — Hakim, Jun 25 '14 at 13:49
Is it $-2\pi \le \theta \le 2\pi$ or $-2\pi \le \theta \le \theta$ ? — Kushashwa Ravi Shrimali, Jun 25 '14 at 13:52

Hakim · Answer 1 · 2014-06-25T14:47:55.967

0

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

edited Jun 25 '14 at 14:47

answered Jun 25 '14 at 14:32

Hakim

10,213

If $\alpha$ is the reference angle of $\theta$, the trigonmetric functions of $\alpha$ should be the same as those of $\theta$ except (possibly) that the sign will change (plus to minus, or minus to plus). This most surely does not happen in most cases when you simply add $\frac{\pi}{2}$ to an angle. – David K Jun 25 '14 at 14:41
@DavidK Thanks, now I understand what is meant by that. Look at my edit. – Hakim Jun 25 '14 at 14:47
OK, that fixes the answer. – David K Jun 25 '14 at 15:05

score 0 · Accepted Answer · answered Jun 25 '14 at 15:03

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?

Trigonometry reference angle of radian

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