Q:

six trigonometric functions of (6,8)

Accepted Solution

A:
[tex]\bf (\stackrel{a}{6}~~,~~\stackrel{b}{8})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+8^2}\implies c=\sqrt{100}\implies c=10\\\\ -------------------------------[/tex]

[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad\qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \\\\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\ -------------------------------[/tex]

[tex]\bf sin(\theta )=\cfrac{8}{10}\implies \cfrac{4}{5}\qquad \qquad cos(\theta )=\cfrac{6}{10}\implies \cfrac{3}{5} \\\\\\ tan(\theta )=\cfrac{8}{6}\implies \cfrac{4}{3}\qquad \qquad cot(\theta )=\cfrac{6}{8}\implies \cfrac{3}{4} \\\\\\ csc(\theta )=\cfrac{10}{8}\implies \cfrac{5}{4}\qquad \qquad sec(\theta )=\cfrac{10}{6}\implies \cfrac{5}{3}[/tex]