We can use following formula
[tex]tan^{2}(A) = \frac{1- cos(2A)}{1+cos(2A)}
\\ \\ tan(A) =2, so
\\ \\4= \frac{1- cos(2A)}{1+cos(2A)}
\\ \\1-cos(2A)=4+4cos(2A)
\\ \\ 5cos(2A) = -3
\\ \\ cos (2A) = -3/5 [/tex]
Now, we need to find sin(2A).
[tex]sin^{2}(2A) + cos^{2}(2A) = 1
\\ \\ sin^{2}(2A) + (- \frac{3}{5})^{2} =1
\\ \\ sin^{2}(2A) =1 - \frac{9}{25}
\\ \\ sin^{2}(2A) = \frac{16}{25}
\\ \\ sin(2A) = \frac{4}{5} \ or \ - \frac{4}{5} [/tex]
Because, the angle A is in the 1st quadrant, and cos(2A) is negative, that means that the angle 2A is in the 2nd quadrant and sin(2A) is a positive number,so sin(2A) = 4/5.
sin(2A) = 4/5
cos(2A) = -3/5
