Help

deleted

1) \(sinθ=\frac{1}{6}\)

and  \(tanθ=\frac{sinθ}{cosθ}\) <=> \(cosθ=\frac{sinθ}{tanθ}=\frac{\frac{1}{6}}{\frac{-\sqrt{35}}{35}}=\frac{-\sqrt{35}}{6 }\) 

so \(cosθ= \frac{-\sqrt{35}}{6}\)

2) \(cosθ=cos(-θ)=\frac{\sqrt{3}}{4}\)

and \(cos^2θ + sin^2θ =1 \) 

so  \(sinθ =\sqrt{1-cos^2θ} = \sqrt{\frac{16}{16}-\frac{3}{16}}=\sqrt{\frac{13}{16}}=\frac{\sqrt{13}}{-4}\)

so \(sinθ=-\frac{\sqrt{13}}{4}\)

Hope this helps!