sry plz proof me

I do not thinkt that this is true  

Essentially this is the question

prove

$$\dfrac{v^2sin^2\alpha}{2g}=\dfrac{-g}{2}\left(v^2\times\dfrac{sin\alpha}{g}\right)^2+(vsin\alpha)\left(\dfrac{vsin\alpha}{g}\right)\\\\\\
\begin{array}{rll}
RHS&=&\frac{-g}{2}\left(v^4\times\frac{sin^2\alpha}{g^2}\right)+\left(\frac{v^2sin^2\alpha}{g}\right)\\\\
&=&\frac{-1}{2}\left(\frac{v^4sin^2\alpha}{g}\right)+\left(\frac{v^2sin^2\alpha}{g}\right)\\\\
&=&\left(\frac{-v^4sin^2\alpha}{2g}\right)+\left(\frac{2v^2sin^2\alpha}{2g}\right)\\\\
&=&\frac{v^2sin^2\alpha}{2g}\;(-v^2+2)\\\\
&=&\frac{v^2sin^2\alpha\;(2-v^2)}{2g}\;\\\\
\end{array}\\\\
$ This does not equal the LHS $$$