Question
Find the trigonometric functions of : – 60°
✓
Answer
Angle of measure $\left(-60^{\circ}\right)$ :
Let $m_{\angle} X O A=-60^{\circ}$
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. $O P=1$,
$-\frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{2}$
Since point $P$ lies in the 4 quadrant,
$ x>0, y<0$
$x=O M=\frac{1}{2} \text { and } y=-P M=-\frac{\sqrt{7}}{2}$
$\therefore \quad \mathbf{P}=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
$\sin \left(-60^{\circ}\right)-y=-\frac{\sqrt{3}}{2}$
$\cos \left(-60^{\circ}\right)-x=\frac{1}{2}$
$\tan \left(-60^{\circ}\right)=\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
$=-\sqrt{3}$
$\operatorname{cosec}\left(-60^{\circ}\right)-\frac{1}{y}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}$
$\sec \left(-60^{\circ}\right)=\frac{1}{x} =\frac{1}{\left(\frac{1}{2}\right)}=2$
$\cot \left(-60^{\circ}\right)=\frac{x}{\sqrt{3}} =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$ =-\frac{1}{\sqrt{3}}$
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