Let us consider 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
\[sin\text{ }5x\text{ }=\text{ }\left( sin\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }cos\text{ }2x\text{ }sin\text{ }x\text{ } \right)\text{ }cos\text{ }2x\text{ }+\text{ }\left( \text{ }cos\text{ }2x\text{ }cos\text{ }x\text{ }\text{ }sin\text{ }2x\text{ }sin\text{ }x \right)\text{ }sin\text{ }2x\]… (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
\[sin\text{ }5x\text{ }=\text{ }[\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }cos\text{ }x\text{ }+\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }sin\text{ }x]\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }+\text{ }[(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }cos\text{ }x\text{ }\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }sin\text{ }x)]\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\]
\[=\text{ }[2\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }+\text{ }sin\text{ }x\text{ }co{{s}^{2}}x\text{ }\text{ }si{{n}^{3}}x]\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }+\text{ }[co{{s}^{3}}x\text{ }\text{ }si{{n}^{2}}x\text{ }cos\text{ }x\text{ }\text{ }2\text{ }si{{n}^{2}}~x\text{ }cos\text{ }x]\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\]
\[=\text{ }co{{s}^{2}}x\text{ }[3\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{3}}x]\text{ }\text{ }si{{n}^{2}}x\text{ }[3\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{3}}x]\text{ }+\text{ }2\text{ }sin\text{ }x\text{ }co{{s}^{4}}x\text{ }\text{ }2\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }4\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\]
\[=\text{ }3\text{ }sin\text{ }x\text{ }co{{s}^{4}}~x\text{ }\text{ }si{{n}^{3}}x\text{ }co{{s}^{2}}x\text{ }\text{ }3\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{5}}x\text{ }+\text{ }2\text{ }sin\text{ }x\text{ }co{{s}^{4}}x\text{ }\text{ }2\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }4\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\]\[=\text{ }5\text{ }sin\text{ }x\text{ }co{{s}^{4}}~x\text{ }10si{{n}^{3}}xco{{s}^{2}}x\text{ }+si{{n}^{5}}x\]
= RHS
Hence proved.