Question
Find the trigonometric functions of : – 315°
✓
Answer
Angle of measure $\left(315^{\circ}\right)$ :
Let $m \angle X O A 315^{\circ}$
Its terminal arm (ray $O A)$ intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the X-axis.
$\triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$\begin{aligned}
O P= & 1, \\
& O M=\frac{1}{\sqrt{2}} O P=\frac{1}{\sqrt{2}}(1)=\frac{1}{\sqrt{2}} \\
& P M=\frac{1}{\sqrt{2}} O P=\frac{1}{\sqrt{2}}(1)=\frac{1}{\sqrt{2}}
\end{aligned}$
Since point $P$ lies in the $1^{\text {st }}$ quadrant,
$x>0, y>0$
$\therefore \quad x=\mathrm{OM}=\frac{1}{\sqrt{2}} \text { and } y=\mathrm{PM}=\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\sin \left(-315^{\circ}\right)=y$
$=\frac{1}{\sqrt{2}}$
$\cos \left(-315^{\circ}\right)=x$
$=\frac{1}{\sqrt{2}}$
$\tan \left(-315^{\circ}\right)=\frac{y}{x}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$
$\operatorname{cosec}\left(-315^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\sec \left(-315^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot \left(-315^{\circ}\right)=\frac{x}{y}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$
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