Question
Find the trigonometric functions of : 45°

Answer

Image
Angle of measure $45^{\circ}$ :
Let $m \angle X O A=45^{\circ}$
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.
$ \mathrm{OP}  =1,$
$\mathrm{OM}  =\frac{1}{\sqrt{2}} \mathrm{OP}$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
$\mathrm{PM}  =\frac{1}{\sqrt{2}} \mathrm{OP}$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}} $
Since point $P$ lies in the 1 st quadrant, $x>0, y>0$
$\therefore \mathrm{x}=\mathrm{OM}=\frac{1}{\sqrt{2}}$ and
$ y=P M=\frac{1}{\sqrt{2}}$
$\therefore P=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $
$\therefore \mathrm{P}=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\sin 45^{\circ}=y=\frac{1}{\sqrt{2}}$
$\cos 45^{\circ}=x=\frac{1}{\sqrt{2}}$
$\tan 45^{\circ}=\frac{y}{x}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$
$\operatorname{cosec} 45^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\sec 45^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot 45^{\circ}=\frac{x}{y}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$

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