Question
Find the trigonometric functions of : – 240°
✓
Answer
Angle of measure (240): Let $m \angle X O A=240^{\circ}$ Its terminal arm (ray $\mathrm{OA}$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis. $\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-900$ triangle. $\begin{aligned} \mathrm{OP} & =1 \\ \mathrm{PM} & =\frac{\sqrt{3}}{2} \mathrm{OP} \\ & =\frac{\sqrt{3}}{2}(1) \\ & =\frac{\sqrt{3}}{2} \\ \mathrm{OM} & =\frac{1}{2} \mathrm{OP} \\ & =\frac{1}{2}(1)=\frac{1}{2} \end{aligned}$ Since point $P$ lies in the 2 nd quadrant, $x<0, y>0$ $\begin{array}{c} \therefore \quad x=-\mathrm{OM}=-\frac{1}{2} \text { and } y=\mathrm{PM}=\frac{\sqrt{3}}{2} \\ \therefore \quad \mathrm{P} \equiv\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \\ \sin \left(-240^{\circ}\right)=y=\frac{\sqrt{3}}{2} \\ \cos \left(-240^{\circ}\right)=x=-\frac{1}{2} \\ \tan \left(-240^{\circ}\right)=\frac{y}{x}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3} \\ \operatorname{cosec}\left(-240^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}} \\ \sec \left(-240^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2 \\ \cot \left(-240^{\circ}\right)=\frac{x}{y}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}} \end{array}$
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