• sin 2x = -12/13
,• cos 2x = -5/13
,• tan 2x = 12/5
Explanation:Given that
[tex]\tan x=-\frac{3}{2}[/tex]Then
[tex]\begin{gathered} \sin2x=\frac{2\tan x}{1+\tan^2x} \\ \\ =\frac{2(-\frac{3}{2})}{1+(-\frac{3}{2})^2}=\frac{-3}{\frac{13}{4}} \\ \\ =-3\times\frac{4}{13}=-\frac{12}{13} \end{gathered}[/tex][tex]\begin{gathered} \cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac{1-(-\frac{3}{2})^2}{1+(-\frac{3}{2})^2} \\ \\ =\frac{1-\frac{9}{4}}{1+\frac{9}{4}}=\frac{-\frac{5}{4}}{\frac{13}{4}}=-\frac{5}{4}\times\frac{4}{13}=-\frac{5}{13} \end{gathered}[/tex][tex]\begin{gathered} \tan2x=\frac{2\tan x}{1-\tan^2x}=\frac{2(-\frac{3}{2})}{1-(-\frac{3}{2})^2} \\ \\ =\frac{-3}{1-\frac{9}{4}}=\frac{-3}{-\frac{5}{4}}=-3\times\frac{-4}{5}=\frac{12}{5} \end{gathered}[/tex]