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