Question
Find the trigonometric functions of $: – 45^\circ$
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Answer
Angle of measure $45^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=45^{\circ}$
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 $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$O P=1$,
Since point $P$ lies in the 4 th quadrant $x>0, y<0$
$ \therefore \quad x=\mathrm{OM}=\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=\frac{-1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$
$\sin \left(-45^{\circ}\right)=y=-\frac{1}{\sqrt{2}}$
$\cos \left(-45^{\circ}\right)=x=\frac{1}{\sqrt{2}}$
$\tan \left(-45^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}}\right)}=-1$
$\operatorname{cosec}\left(-45^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$\sec \left(-45^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot \left(-45^{\circ}\right)=\frac{x}{y}=\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(-\frac{1}{\sqrt{2}}\right)}=-1 $
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