step 1
Find out sine of theta
Remember that
If θ lies in Quadrant IV
then
the value of sine is negative
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]substitute the value of cosine
[tex]\sin ^2\theta+(\frac{8}{17})^2=1[/tex][tex]\sin ^2\theta=1-\frac{64}{289}[/tex][tex]\begin{gathered} \sin ^2\theta=\frac{225}{289} \\ \sin ^{}\theta=-\frac{15}{17} \end{gathered}[/tex]step 2
we know that
[tex]\sin 2\theta=2\sin \theta\cdot\cos \theta[/tex]substitute given values
[tex]\sin 2\theta=2\cdot(-\frac{15}{17})\cdot(\frac{8}{17)}[/tex][tex]\sin 2\theta=-\frac{240}{289}[/tex]Simplify
