I am trying to show that $$ sin(5\pi t) = \frac 12 e^{-j \frac {\pi}2}e^{j5\pi t} - \frac 12 e^{-j \frac {\pi}2}e^{-j5\pi t} $$
I am aware that $$ sin(\theta) = \frac {e^{j\theta} - e^{-j\theta}}{j2} $$
But if you multiply the top and bottom by $j$, then the answer becomes $$ sin(5\pi t) = -\frac 12 je^{5\pi t} + \frac 12 je^{-j5\pi t} $$
I'm really no sure of where to go from here. Any help is appreciated.
On
What you have run into is Euler's formula for complex exponents:
$$e^{j\theta}=\cos(\theta)+j\sin(\theta)$$
Using the identity that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, we get:
$$e^{-j\theta}=\cos(\theta)-i\sin(\theta)$$
Subtract the two equations, and we get:
$$e^{j\theta}-e^{-j\theta}=2j\sin(\theta)\tag1$$
$$\sin(\theta)=\frac{e^{j\theta}-e^{-j\theta}}{2j}$$
Which is what you started with.
But at step $(1)$, we arrive with your derivation. And so, that is the explanation I give to you.
Using Euler's identity $e^{ix}=\cos x+i\sin x$
$$e^{j\pi/2}=\cos\dfrac\pi2+j\sin\dfrac\pi2=j$$
$$\implies e^{-j\pi/2}=j^{-1}=-j$$ as $j^2=-1$