Ezekiel Schroeder

2023-03-11

How to find the six trigonometric functions of 390 degrees?

zarizaraje1

Given:
$\mathrm{sin}390=\mathrm{sin}\left(30+360\right)=\mathrm{sin}30=\frac{1}{2}$
$\mathrm{cos}390=\mathrm{cos}\left(30+360\right)=\mathrm{cos}30=\frac{\sqrt{3}}{2}$
$\mathrm{tan}390=\mathrm{tan}30=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$
$\mathrm{cot}390=\frac{1}{\mathrm{tan}30}=\frac{\sqrt{3}}{1}$
$\mathrm{sec}390=\frac{1}{\mathrm{cos}30}=\frac{2}{\sqrt{3}}$
$\mathrm{csc}390=\frac{1}{\mathrm{sin}30}=\frac{2}{1}$

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