Question 1029439
Use 
{{{sin^2(theta)+cos^2(theta)=1}}}
to solve for {{{sin(theta)}}}
{{{sin^2(theta)+(9/25)=1}}}
{{{sin^2(theta)=16/25}}}}
{{{sin(theta)=0 +- 4/5}}}
Since it's in Q4, {{{sin(theta)<0}}}
{{{sin(theta)=-4/5}}}
So then,
{{{sin(2*theta)=2sin(theta)cos(theta)}}}
{{{sin(2*theta)=2(-4/5)(3/5)}}}
{{{sin(2*theta)=-24/25}}}