Answer
$\displaystyle \frac{119}{169}$
Work Step by Step
Construct a right triangle in which $\sin t= \displaystyle \frac{5}{13}\qquad (\frac{opp.}{hyp.})$.
See image below.
$t=\displaystyle \sin^{-1}\frac{5}{13}.$
Find a using the Pythagorean th.: $\left[\begin{array}{l}
a^{2}+5^{2}=13^{2}\\
a=\sqrt{169-25}\\
a=12
\end{array}\right]$
Now, $\displaystyle \cos t=\frac{adj.}{opp.}=\frac{12}{13}$
Applying one of the trigonometric identities for double angles,
$\displaystyle \cos(2t)=\cos^{2}t-\sin^{2}t=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$
