Let us consider LHS:
\[si{{n}^{2}}~2\pi /5\text{ }\text{ }si{{n}^{2}}~\pi /3\text{ }=\text{ }si{{n}^{2}}~\left( \pi /2\text{ }\text{ }\pi /10 \right)\text{ }\text{ }si{{n}^{2}}~\pi /3\]
we know, \[sin\text{ }\left( 90{}^\circ \text{ }A \right)\text{ }=\text{ }cos\text{ }A\]
So, \[si{{n}^{2}}~\left( \pi /2\text{ }\text{ }\pi /10 \right)\text{ }=\text{ }co{{s}^{2}}~\pi /10\]
\[Sin\text{ }\pi /3\text{ }=\text{ }\surd 3/2\]
Then the above equation becomes,
\[=\text{ }Co{{s}^{2}}~\pi /10\text{ }\text{ }{{\left( \surd 3/2 \right)}^{2}}\]
We know, \[cos\text{ }\pi /10\text{ }=\text{ }\surd \left( 10+2\surd 5 \right)/4\]
the above equation becomes,
= \[{{\left[ \surd \left( 10+2\surd 5 \right)/4 \right]}^{2}}~\text{ }3/4\]
\[=\text{ }\left[ 10\text{ }+\text{ }2\surd 5 \right]/16\text{ }\text{ }3/4\]
\[=\text{ }\left[ 10\text{ }+\text{ }2\surd 5\text{ }\text{ }12 \right]/16\]
\[=\text{ }\left[ 2\surd 5\text{ }\text{ }2 \right]/16\]
\[=\text{ }\left[ \surd 5\text{ }\text{ }1 \right]/8\]
= RHS
Hence proved.