MATH SOLVE

4 months ago

Q:
# Given sin0=6/11 and sec 0<0 find cos0 and tan0 (picture provided)

Accepted Solution

A:

Answer:[tex]cos(0\°)=-\frac{\sqrt{85}}{11}[/tex][tex]tan(0\°) = -\frac{6}{\sqrt{85}}\\\\[/tex]Step-by-step explanation:We know by definition that:[tex]cos ^ 2(x) = 1-sin ^2(x)[/tex]We also know that:[tex]tan(x) = \frac{sin(x)}{cos(x)}[/tex][tex]sec(x) = \frac{1}{cos(x)}[/tex]Then we can use these identities to solve the problemif [tex]sin(0\°) = \frac{6}{11}[/tex] then [tex]cos^2(0\°) = 1-(\frac{6}{11})^2[/tex][tex]cos(0\°) = \±\sqrt{\frac{85}{121}}\\\\cos(0\°)=\±\frac{\sqrt{85}}{11}[/tex]As [tex]sec(x) <0[/tex] then [tex]cos(x) <0[/tex]. Therefore we take the negative root:[tex]cos(0\°)=-\frac{\sqrt{85}}{11}[/tex]Now that we know [tex]sin(0\°)[/tex] and [tex]cos(0\°)[/tex] we can find [tex]tan(0\°)[/tex][tex]tan(0\°) = \frac{sin(0\°)}{cos(0\°)}\\\\tan(0\°) = \frac{\frac{6}{11}}{\frac{-\sqrt{85}}{11} }\\\\tan(0\°) = -\frac{6}{\sqrt{85}}\\\\[/tex]