Find the trigonometric functions of :

Question

Find the trigonometric functions of : – 120°

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Answer

Angle of measure $\left(-120^{\circ}\right)$ :
Let $m \angle X O A=-120^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$,
Since point $P$ lies in the 3rd quadrant, $x<0, y<0$
$\begin{array}{c}
\therefore \quad x=-\mathrm{OM}=\frac{-1}{2} \text { and } y=-\mathrm{PM}=\frac{-\sqrt{3}}{2} \\
\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2}\right)\\
\sin \left(-120^{\circ}\right)=y=-\frac{\sqrt{3}}{2} \\
\cos \left(-120^{\circ}\right)=x=-\frac{1}{2} \\
\tan \left(-120^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3} \\
\operatorname{cosec}\left(-120^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}} \\
\sec \left(-120^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2 \\
\cot \left(-120^{\circ}\right)=\frac{x}{y}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}
\end{array}$