Question
Find the trigonometric functions of $: – 30^\circ$
✓
Answer
Angle of measure $30^{\circ}$
Let $m \angle X O A=-30^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60-90^{\circ}$ triangle.
$ \mathrm{Op} =1,$
$\mathrm{OP} =1,$
$\mathrm{OM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2}$
$\mathrm{PM} =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)=\frac{1}{2} $
Since point $P$ lies in the 4 th quadrant $x>0, y<0$
$ \therefore \quad x=\mathrm{OM}=\frac{\sqrt{3}}{2} \text { and } y=-\mathrm{PM}=\frac{-1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{-1}{2}\right)$
$\sin \left(-30^{\circ}\right)=y=-\frac{1}{2}$
$\cos \left(-30^{\circ}\right)=x=\frac{\sqrt{3}}{2}$
$\tan \left(-30^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=-\frac{1}{\sqrt{3}}$
$\operatorname{cosec}\left(-30^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sec \left(-30^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot \left(-30^{\circ}\right)=\frac{x}{y}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=-\sqrt{3} $
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