Respuesta :

[tex]\bf sin(-\theta)\iff -sin(\theta)\qquad \qquad tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\implies cos(\theta)=\cfrac{sin(\theta)}{tan(\theta)}\quad \begin{cases} sin(-\theta)=\cfrac{1}{5}\\\\ -sin(\theta)=\cfrac{1}{5}\\\\ sin(\theta)=-\cfrac{1}{5}\\ --------\\ tan(\theta)=\cfrac{\sqrt{6}}{12} \end{cases} \\\\\\ thus\qquad cos(\theta)=\cfrac{-\frac{1}{5}}{\frac{\sqrt{6}}{12}}[/tex]

simplify away

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