Answer
$$e^{-t+\frac{\pi}{6}} [\dfrac{\sin 2t}{2}-\dfrac{\sqrt 3}{2} \cos (2t)]$$
Work Step by Step
We are given that $f(t)=e^{-t} \sin (2t)$ and $a=\dfrac{\pi}{6}$
When $a$ has a positive value then , shift the function to the right for $a$ units and when $a$ has a negative value then , shift the function to the left for $a$ units
Now, $f(t-a) =f(t-\dfrac{\pi}{6}) \\=e^{-t+\frac{\pi}{6}} \sin 2(t-\dfrac{\pi}{6}) \\=e^{-t+\frac{\pi}{6}} [\dfrac{\sin 2t}{2}-\dfrac{\sqrt 3}{2} \cos (2t)] $
