Question
Find the trigonometric functions of : 315°
✓
Answer
Angle of measure $315^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=315^{\circ}$
Its terminal arm (ray $O A)$ intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
Since point $P$ lies in the $4^{\text {th }}$ quadrant,
$x>0, y<0$
$\therefore \quad x=\mathrm{OM}=\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=-\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$\sin 315^{\circ}=y=-\frac{1}{\sqrt{2}}$
$\cos 315^{\circ}=x=\frac{1}{\sqrt{2}}$
$\tan 315^{\circ}=\frac{y}{x}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$
$\operatorname{cosec} 315^{\circ}=\frac{1}{y}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$ \sec 315^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot 315^{\circ}=\frac{x}{\dot{y}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1 $
[Note: Answer given in the textbook of $\cot 315^{\circ}$ is 1. However, as per our calculation it is -1.]
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