Let us consider LHS:
\[cos\text{ }{{78}^{o}}~cos\text{ }{{42}^{o}}~cos\text{ }{{36}^{o}}\]
Let us multiply and divide by 2 we get,
\[cos\text{ }{{78}^{o}}~cos\text{ }{{42}^{o}}~cos\text{ }{{36}^{o}}~=\text{ }1/2\text{ }(2\text{ }cos\text{ }{{78}^{o}}~cos\text{ }{{42}^{o}}~cos\text{ }{{36}^{o}})\]
We know, \[2\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }=\text{ }cos\text{ }\left( A\text{ }+\text{ }B \right)\text{ }+\text{ }cos\text{ }\left( A\text{ }\text{ }B \right)\]
Then the above equation becomes,
\[=\text{ }1/2\text{ }(cos\text{ }({{78}^{o}}~+\text{ }{{42}^{o}})\text{ }+\text{ }cos\text{ }({{78}^{o}}~\text{ }{{42}^{o}}))\text{ }\times \text{ }cos\text{ }{{36}^{o}}\]
\[=\text{ }1/2\text{ }(cos\text{ }{{120}^{o}}~+\text{ }cos\text{ }{{36}^{o}})\text{ }\times \text{ }cos\text{ }{{36}^{o}}\]
\[=\text{ }1/2\text{ }(cos\text{ }({{180}^{o}}~\text{ }{{60}^{o}})\text{ }+\text{ }cos\text{ }{{36}^{o}})\text{ }\times \text{ }cos\text{ }{{36}^{o}}\]
\[=\text{ }1/2\text{ }(-cos\text{ }({{60}^{o}})\text{ }+\text{ }cos\text{ }{{36}^{o}})\text{ }\times \text{ }cos\text{ }{{36}^{o}}~\left[ since,\text{ }cos\left( 180{}^\circ \text{ }\text{ }A \right)\text{ }=\text{ }\text{ }A \right]\]
\[=\text{ }1/2\text{ }\left( -1/2\text{ }+\text{ }\left( \surd 5\text{ }+\text{ }1 \right)/4 \right)\text{ }\left( \left( \surd 5\text{ }+\text{ }1 \right)/4 \right)\text{ }[since,\text{ }cos\text{ }{{36}^{o}}~=\text{ }\left( \surd 5\text{ }+\text{ }1 \right)/4]\]
\[=\text{ }1/2\text{ }\left( \surd 5\text{ }+\text{ }1\text{ }\text{ }2 \right)/4\text{ }\left( \left( \surd 5\text{ }+\text{ }1 \right)/4 \right)\]
\[=\text{ }1/2\text{ }\left( \surd 5\text{ }\text{ }1 \right)/4)\text{ }\left( \left( \surd 5\text{ }+\text{ }1 \right)/4 \right)\]
\[=\text{ }1/2\text{ }({{\left( \surd 5 \right)}^{2}}~\text{ }{{1}^{2}})/16\]
\[=\text{ }1/2\text{ }\left( 5-1 \right)/16\]
\[=\text{ }1/2\text{ }\left( 4/16 \right)\]
\[=\text{ }1/8\]
= RHS
Hence proved.