Question
Find the trigonometric functions of $: – 225^\circ$
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Answer
Angle of measure $\left(-225^{\circ}\right)$ : Let $m \angle X O A=-225^{\circ}$ 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 $45^{\circ}-45^{\circ}-90^{\circ}$ triangle. $\mathrm{OP}=1$, Since point $P$ lies in the $2$ nd quadrant, $x<0, y>0$
$ \therefore \quad x=-\mathrm{OM}=\frac{-1}{\sqrt{2}} \text { and } y=\mathrm{PM}=\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\sin \left(-225^{\circ}\right)=y=\frac{1}{\sqrt{2}}$
$\cos \left(-225^{\circ}\right)=x=-\frac{1}{\sqrt{2}}$
$\tan \left(-225^{\circ}\right)=\frac{y}{x}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1$
$\operatorname{cosec}\left(-225^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\sec \left(-225^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$\cot \left(-225^{\circ}\right)=\frac{x}{y}=\frac{1}{\frac{1}{\sqrt{2}}}=-1$
$\frac{\sqrt{2}}{} $
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