The first step to answer this question is to find tan(2u) by using the double angle formula:
[tex]\begin{gathered} tan(2u)=\frac{2tan(u)}{1-tan^2(u)} \\ tan(2u)=\frac{2(\frac{13}{5})}{1-(\frac{13}{5})^2} \\ tan(2u)=\frac{\frac{26}{5}}{1-\frac{169}{25}} \\ tan(2u)=\frac{\frac{26}{5}}{-\frac{144}{25}} \\ tan(2u)=-\frac{65}{72} \end{gathered}[/tex]It means that tan(2u) is -65/72.
The next step is to rewrite the equations for sin(2u) and cos(2u) to have them in terms of the least number of variables possible, this way:
[tex]\begin{gathered} sin(2u)=2sin(u)cos(u) \\ sin(2u)=2sin(u)\frac{sin(u)}{tan(u)} \\ sin(2u)=\frac{2sin^2(u)}{tan(u)} \end{gathered}[/tex][tex]\begin{gathered} cos(2u)=cos{}^2(u)-sin^2(u) \\ cos(2u)=1-sin^2(u)-s\imaginaryI n^2(u) \\ cos(2u)=1-2sin^2(u) \end{gathered}[/tex]If we rewrite tan(2u) in terms of sin(2u) and cos(2u) we will have:
[tex]\begin{gathered} tan(2u)=\frac{sin(2u)}{cos(2u)} \\ tan(2u)=\frac{\frac{2s\imaginaryI n^{2}(u)}{tan(u)}}{1-2sin^2(u)} \end{gathered}[/tex]We know the values of tan(2u) and tan(u), so we can solve the equation for sin^2(u).
[tex]\begin{gathered} tan(2u)=\frac{2sin^2(u)}{tan(u)(1-2sin^2(u))} \\ -\frac{65}{72}=\frac{2s\imaginaryI n^2(u)}{\frac{13}{5}(1-2s\imaginaryI n^2(u))} \\ -\frac{65}{72}\cdot\frac{13}{5}\cdot(1-2sin^2(u))=2sin^2(u) \\ -\frac{169}{72}(1-2sin^2(u))=2sin^2(u) \\ -1+2sin^2(u)=\frac{72}{169}\cdot2sin^2(u) \\ -1+2sin^2(u)=\frac{144}{169}sin^2(u) \\ 2sin^2(u)-\frac{144}{169}sin^2(u)=1 \\ \frac{194}{169}sin^2(u)=1 \\ sin^2(u)=\frac{169}{194} \end{gathered}[/tex]Using this value we can find the values of sin(2u) and cos(2u):
[tex]\begin{gathered} sin(2u)=\frac{2sin^2(u)}{tan(u)} \\ sin(2u)=\frac{2\cdot\frac{169}{194}}{\frac{13}{5}} \\ sin(2u)=\frac{65}{97} \end{gathered}[/tex][tex]\begin{gathered} cos(2u)=1-2sin^2(u) \\ cos(2u)=1-2\cdot\frac{169}{194} \\ cos(2u)=1-\frac{169}{97} \\ cos(2u)=-\frac{72}{97} \end{gathered}[/tex]It means that sin(2u)=65/97, cos(2u)=-72/97 and tan(2u)=-65/72.
