Question
Find the trigonometric functions of : -300°
✓
Answer
Angle of measure $\left(-300^{\circ}\right)$ :
Let $m \angle X O A=-300^{\circ}$ Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the $X$-axis.
$\triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$
$ \mathrm{OM} =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2}$
$\mathrm{PM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$=\frac{\sqrt{3}}{2} $
Since point $P$ lies in the 1st quadrant, $x>0, y>0$ $ x=O M=\frac{1}{2} \text { and }$
$y=P M=\frac{\sqrt{3}}{2} $
$\therefore \quad P=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$ \sin \left(-300^{\circ}\right)=y=\frac{\sqrt{3}}{2}$
$\cos \left(-300^{\circ}\right)=x=\frac{1}{2} $
$\tan \left(-300^{\circ}\right)=\frac{y}{x}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2}\right)}=\sqrt{3}$
$\operatorname{cosec}\left(-300^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$ \sec \left(-300^{\circ}\right)=\frac{1}{x}=\left(\frac{1}{\frac{1}{2}}\right)=2$
$\cot \left(-300^{\circ}\right)=\frac{x}{y}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}} $
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