Take LHS:

\[sin\text{ }4x\]

From formula, \[sin\text{ 2}x\text{ }=\text{ }2\text{ }sin\text{ }x\text{ }cos\text{ }x\]

\[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\]

So,

\[sin\text{ }4x\text{ }=\text{ }2\text{ }sin\text{ }2x\text{ }cos\text{ }2x\]

\[=\text{ }2\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)\]

\[=\text{ }4\text{ }sin\text{ }x\text{ }cos\text{ }x\text{ }(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)\]

\[=\text{ }4\text{ }sin\text{ }x\text{ }co{{s}^{3}}~x\text{ }\text{ }4\text{ }si{{n}^{3}}~x\text{ }cos\text{ }x\]

= RHS

Hence proved.