Question
Prove that:
$\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
$\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
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$\frac{x+i y}{2+3 i}+\frac{2+i}{2-3 i}=\frac{9}{13}(1+i)$