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.