Respuesta :

[tex]\bf \textit{symmetry identity}\qquad sin(-\theta)\iff -sin(\theta)\\\\ -----------------------------\\\\ sin(-\theta)=\cfrac{1}{5}\implies -sin(\theta)=\cfrac{1}{5}\implies sin(\theta)=-\cfrac{1}{5} \\\\\\ tan(\theta)=\cfrac{\sqrt{6}}{12}\\\\ -----------------------------\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\implies cos(\theta)=\cfrac{sin(\theta)}{tan(\theta)}\implies cos(\theta)=\cfrac{-\frac{1}{5}}{\frac{\sqrt{6}}{12}}[/tex]

simplify away
ghanami
hello : 
tan(θ)= Sin(θ)/cos(θ) 

cos(θ) =Sin(θ)/tan(θ)
Sin(−θ)= - Sin(θ)
Sin(θ) = -1/5
cos(θ) =(1/5)/√6/12

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