We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
133
4
avatar

1.

2.

 Nov 12, 2018
 #1
avatar+18133 
0

deleted

 Nov 12, 2018
edited by Guest  Nov 12, 2018
edited by Guest  Nov 12, 2018
 #2
avatar+316 
+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+316 
+1

Why? Where is my false? 

Dimitristhym  Nov 12, 2018

28 Online Users

avatar
avatar
avatar