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2.

 Nov 12, 2018
 #1
avatar+20440 
0

deleted

 Nov 12, 2018
edited by Guest  Nov 12, 2018
edited by Guest  Nov 12, 2018
 #2
avatar+342 
+2

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! 

 Nov 12, 2018
edited by Dimitristhym  Nov 12, 2018
edited by Dimitristhym  Nov 12, 2018
 #3
avatar
0

^^ thats not correct

the correct answer was -√35 / 6

and - √13 / 4

Guest Nov 12, 2018
edited by Guest  Nov 12, 2018
 #4
avatar+342 
+1

Why? Where is my false? 

Dimitristhym  Nov 12, 2018

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