Take LHS:

\[sin\text{ }5x\]

We can write \[sin\text{ }5x\] as,

\[sin\text{ }5x\text{ }=\text{ }sin\text{ }\left( 3x\text{ }+\text{ }2x \right)\]

From formula,

\[Sin\text{ }\left( x\text{ }+\text{ }y \right)\text{ }=\text{ }sin\text{ }x\text{ }cos\text{ }y\text{ }+\text{ }cos\text{ }x\text{ }sin\text{ }y\ldots ..\left( i \right)\]

So,

\[sin\text{ }5x\text{ }=\text{ }sin\text{ }3x\text{ }cos\text{ }2x\text{ }+\text{ }cos\text{ }3x\text{ }sin\text{ }2x\]

\[=\text{ }sin\text{ }\left( 2x\text{ }+\text{ }x \right)\text{ }cos\text{ }2x\text{ }+\text{ }cos\text{ }\left( 2x\text{ }+\text{ }x \right)\text{ }sin\text{ }2x\] ……..(ii)

And

cos (x + y) = cos x cos y – sin x sin y……(iii)

Now substituting equation (i) and (iii) in equation (ii), we get

\[sin\text{ }5x\text{ }=\text{ }\left( sin\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }cos\text{ }2x\text{ }sin\text{ }x\text{ } \right)\text{ }cos\text{ }2x\text{ }+\text{ }\left( \text{ }cos\text{ }2x\text{ }cos\text{ }x\text{ }\text{ }sin\text{ }2x\text{ }sin\text{ }x \right)\text{ }sin\text{ }2x\]

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

\[=\text{ }2sin\text{ }2x\text{ }cos\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }co{{s}^{2}}~2x\text{ }sin\text{ }x\text{ }\text{ }si{{n}^{2}}~2x\text{ }sin\text{ }x\] …….(iv)

Now \[sin\text{ }2x\text{ }=\text{ }2sin\text{ }x\text{ }cos\text{ }x\] ………(v)

And \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x\] ………(vi)

Substituting equation (v) and (vi) in equation (iv), we get

\[sin\text{ }5x\text{ }=\text{ }2\left( 2sin\text{ }x\text{ }cos\text{ }x \right)\text{ }(co{{s}^{2}}x\text{ }si{{n}^{2}}x)\text{ }cos\text{ }x\text{ }+\text{ }{{(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)}^{2~}}sin\text{ }x\text{ }\text{ }{{\left( 2sin\text{ }x\text{ }cos\text{ }x \right)}^{2~}}sin\text{ }x\]\[=\text{ }4(sin\text{ }x\text{ }co{{s}^{2}}~x)\text{ }([1\text{ }si{{n}^{2}}x]\text{ }\text{ }si{{n}^{2}}x)\text{ }+\text{ }{{([1si{{n}^{2}}x]\text{ }\text{ }si{{n}^{2}}x)}^{2~}}sin\text{ }x\text{ }\text{ }(4si{{n}^{2}}~x\text{ }co{{s}^{2}}~x)sin\text{ }x\] (as \[co{{s}^{2}}x\text{ }+\text{ }si{{n}^{2}}x\text{ }=\text{ }1~\Rightarrow ~co{{s}^{2}}x\text{ }=\text{ }1\text{ }si{{n}^{2}}x\])

\[sin\text{ }5x\text{ }=\text{ }4(sin\text{ }x\text{ }[1\text{ }\text{ }si{{n}^{2}}x])\text{ }(1\text{ }\text{ }2si{{n}^{2}}x)\text{ }+\text{ }{{(1\text{ }\text{ }2si{{n}^{2}}x)}^{2~}}sin\text{ }x\text{ }\text{ }4si{{n}^{3}}~x\text{ }[1\text{ }\text{ }si{{n}^{2}}x]\]

\[=\text{ }4sin\text{ }x\text{ }(1\text{ }\text{ }si{{n}^{2}}x)\text{ }(1\text{ }\text{ }2si{{n}^{2~}}x)\text{ }+\text{ }(1\text{ }\text{ }4si{{n}^{2}}x\text{ }+\text{ }4si{{n}^{4}}x)\text{ }sin\text{ }x\text{ }\text{ }4si{{n}^{3}}~x\text{ }+\text{ }4si{{n}^{5}}x\]

\[=\text{ }(4sin\text{ }x\text{ }\text{ }4si{{n}^{3}}x)\text{ }(1\text{ }\text{ }2si{{n}^{2}}x)\text{ }+\text{ }sin\text{ }x\text{ }\text{ }4si{{n}^{3}}x\text{ }+\text{ }4si{{n}^{5}}x\text{ }\text{ }4si{{n}^{3}}~x\text{ }+\text{ }4si{{n}^{5}}x\]

\[=\text{ }4sin\text{ }x\text{ }\text{ }8si{{n}^{3}}x\text{ }\text{ }4si{{n}^{3}}x\text{ }+\text{ }8si{{n}^{5}}x\text{ }+\text{ }sin\text{ }x\text{ }\text{ }8si{{n}^{3}}x\text{ }+\text{ }8si{{n}^{5}}x\]

\[=\text{ }5sin\text{ }x\text{ }\text{ }20si{{n}^{3}}x\text{ }+\text{ }16si{{n}^{5}}x\]

= RHS

Hence proved.