Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$ 
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
 \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
 \alpha$.

I tried:
$$\frac{\frac{\sin \alpha}{\cos \alpha}+\sin \alpha}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha+\sqrt{5}\sin \alpha \cos \alpha}{\cos \alpha}}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha}{\cos \alpha}(\sqrt{5}\cos \alpha+1)}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\tan \alpha (5\cos^2\alpha-1)}{2} = \\
???$$
What do I do next?

Problem:
$$
f(t) = \frac{1}{2} \left(\sqrt{5} \cos t-1\right) \left(\sqrt{5} \sin t+\tan t\right)
$$
FOIL Components:
$$
%
\begin{align}
%
 F 
&= \frac{1}{2} \sqrt{5}  \cos t
 \left(\sqrt{5} \sin t\right)
= \frac{5}{2} \sin t \cos t\\[7pt]
%
O &= \frac{1}{2} \sqrt{5}  \cos t \left( \tan t\right) 
= \frac{\sqrt{5}}{2}  \sin t \\[7pt]
%
 I &= \frac{1}{2}\left(-1\right) \sqrt{5} \sin t = -\frac{\sqrt{5}}{2} \sin t \\[7pt]
%
 L &= \frac{1}{2}\left(-1\right)\tan t = -\frac{1}{2}\tan t
%
\end{align}
%
$$
Combine and simplify
$$
\begin{align}
\require{cancel}
f(t) &= F + O + I + L \\
&= \frac{5}{2} \sin t \cos t + 
\cancel{\frac{\sqrt{5}}{2}  \sin t} - 
\cancel{\frac{\sqrt{5}}{2}  \sin t} -
\frac{1}{2}\tan t \\
&=
\frac{5}{2} \sin t \cos t - \frac{1}{2}\tan t
\end{align}
$$
Noting
$$
 \sin t \cos t = \frac{1}{2} \sin 2t
$$
The function reduces to the desired form
$$
\boxed{
f(t) = \frac{5}{4} \sin 2t - \frac{1}{2}\tan t
}
$$
