Chaya Galloway

2021-08-14

Use the unit circle to evaluate the six trigonometric functions of: a) ${90}^{\circ }$ b) ${180}^{\circ }$

Arham Warner

Given:
Use the unit circle to evaluate the six trigonometric functions:
a) 90
b) 180
Any point in on the unit circle is given by:
$⇒\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)$
a) 90:
Point on the unit circle is given by:
$⇒\left(\mathrm{cos}90,\mathrm{sin}90\right)$
$⇒\left(0,1\right)$
So,
$⇒\mathrm{sin}90=1$
$⇒\mathrm{cos}90=0$
$⇒\mathrm{tan}90=\frac{\mathrm{sin}90}{\mathrm{cos}90}=\frac{1}{0}=\text{not defined}$
$⇒\mathrm{csc}90=\frac{1}{\mathrm{sin}90}=1$
$⇒\mathrm{sec}90=\frac{1}{\mathrm{cos}90}=\text{not defined}$
$⇒\mathrm{cot}90=\frac{1}{\mathrm{tan}90}=0$
b) 180:
Point on the unit circle is given by:
$⇒\left(\mathrm{cos}180,\mathrm{sin}180\right)$
$⇒\left(-1,0\right)$
So,
$⇒\mathrm{sin}180=0$
$⇒\mathrm{cos}180=-1$
$⇒\mathrm{tan}180=\frac{\mathrm{sin}180}{\mathrm{cos}180}=\frac{0}{-1}=0$
$⇒\mathrm{csc}180=\frac{1}{\mathrm{sin}180}=\text{not defined}$
$⇒\mathrm{sec}180=\frac{1}{\mathrm{cos}180}=-1$
$⇒\mathrm{cot}180=\frac{1}{\mathrm{tan}180}=\text{not defined}$

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