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