Question
Find the trigonometric functions of $: -330^\circ$
✓
Answer
Angle of measure $\left(-330^{\circ}\right)$ :
Let $m \angle X O A=-330^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the $\mathrm{X}$-axis.
$\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$
$ \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 1st quadrant, $x>0, y>0$
$ \therefore \mathrm{x}=\mathrm{OM}=\frac{\sqrt{3}}{2} \text { and } \mathrm{y}=\mathrm{PM}=\frac{1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\sin \left(-330^{\circ}\right)=y=\frac{1}{2}$
$\cos \left(-330^{\circ}\right)=x=\frac{\sqrt{3}}{2}$
$\tan \left(-330^{\circ}\right)=\frac{y}{x}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$
$\operatorname{cosec}\left(-330^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\sec \left(-330^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot \left(-330^{\circ}\right)=\frac{x}{y}=\frac{\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)}=\sqrt{3} $
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