Question 925282
You are right in that
{{{sin^2(t)=8/9}}} and
{{{sin(t)= -2sqrt(2)/3}}} (since {{{sin(t)<0}}} as the problem says)
 
For any {{{t}}} , {{{sin(-t)=-sin(t)}}} , so
{{{highlight(sin(-t)=2sqrt(2)/3)}}}
 
Using trigonometric identity formulas,
you can find {{{cos(t+5pi/3)}}} and {{{sin(2t)}}} :
 
One of the most popular trigonometric identities is
{{{cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B)}}} , so
{{{cos(t+5pi/3)=cos(t)*cos(5pi/3)-sin(t)*sin(5pi/3)}}} ,
and since {{{cos(5pi/3)=cos(5pi/3-2pi)=cos(-pi/3)=cos(pi/3)=1/2}}}
and {{{sin(5pi/3)=sin(5pi/3-2pi)=sin(-pi/3)=-sin(pi/3)=-sqrt(3)/2}}}
{{{cos(t+5pi/3) =(-1/3)*(1/2)-(-2sqrt(2)/3)*(-sqrt(3)/2)}}}
{{{cos(t+5pi/3) =-1/6-(2sqrt(2)sqrt(3)/(2*3))}}}
{{{highlight(cos(t+5pi/3) =-1/6-sqrt(6)/3)}}}
 
An other popular trigonometric identity is
{{{sin(2A)=2sin(A)cos(A)}}} , so
{{{sin(2t)=2sin(t)cos(t)}}}
{{{sin(2t)=2(-2sqrt(2)/3)(-1/3)}}}
{{{highlight(sin(2t)=4sqrt(2)/9)}}}