Question
$\sin5\text{x}=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$

Answer

$\text{LHS}=\sin5\text{x}$
$=\sin\text{x}(3\text{x}+2\text{x})$
$=\sin3\text{x}\times\cos2\text{x}+\cos3\text{x}\times\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2-1)\\+(4\cos^32\cos^2-3\cos\text{x})\times2\sin\text{x}\cos\text{x}$
$=-3\sin\text{x}-4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}\\-8\sin^3\text{x}\cos^2\text{x}+8\sin\text{x}\cos^4\text{x}-6\sin\text{x}\cos^2\text{x}$
$=8\sin\text{x}\cos^4\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\sin\text{x}\cos^4\text{x}​​+4\sin^3​\text{x}​+2\sin^3\text{x}\cos^2​​\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3(1+\cos^2\text{x})-2(2+\cos^2\text{x})]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3+3\cos^2\text{x}-4-2\cos^2\text{x}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}[\cos^2\text{x-1}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}\times(-\sin^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
$=\text{RHS}$

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