Find the exact value of sin(2y) given sin(y)=-8/17 and 270

Find the exact value of sin(2y) given sin(y)=-8/17 and 270<y<360

4th quad so cos y is positive.

\(sin2y=2sinycosy\)

If opp is 8 and hyp is 17 then adjacent must be   

\(adj^2=17^2-8^2=289-64=225\\ adj = 15 \)

\(sin2y=2sinycosy\\ sin2y=2*\frac{-8}{17}*\frac{15}{17}\\ sin2y=2*\frac{-120}{289}\\ sin2y=\frac{-240}{289}\\\)

Nice use of that identity, Melody.....!!!

Thanks Chris     

A note to guest asker:

To get the 15 I would normally draw up a little right anglesd triangle - it is easier to see visually.  :)