Question
Find the trigonometric functions of : 225°
✓
Answer
Angle of measure $225^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=225^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg $\mathrm{PM}$ perpendicular to the $X$-axis.
$\triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$ O P =1$
$O M =\frac{1}{\sqrt{2}} O P$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
$P M =\frac{1}{\sqrt{2}} O P$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$ \therefore \quad x=-\mathrm{OM}=-\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=-\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$\sin 225^{\circ}=y=-\frac{1}{\sqrt{2}}$
$\cos 225^{\circ}=x=-\frac{1}{\sqrt{2}}$
$\tan 225^{\circ}=\frac{y}{x}=\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=1$
$\operatorname{cosec} 225^{\circ}=\frac{1}{y}=\frac{1}{-\frac{1}{\sqrt{2}}}=-\sqrt{2}$
$\sec 225^{\circ}=\frac{1}{x}=\frac{1}{-\frac{1}{\sqrt{2}}}=-\sqrt{2}$
$\cot 225^{\circ}=\frac{x}{y}=\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=1 $
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