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Sin(x) + sin(3x) = 4 sin(x) cos^2(x)
Prove the identity.

User Dabo
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1 Answer

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\bf \textit{Sum to Product Identities} \\\\ sin(\alpha)+sin(\beta)=2sin\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right)\leftarrow \textit{we'll use this one} \\\\\\ sin(\alpha)-sin(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ \stackrel{\textit{symmetry identity}}{cos(-\theta )=cos(\theta )} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf sin(x)+sin(3x)=\underline{4sin(x)cos^2(x)} \\\\[-0.35em] ~\dotfill\\\\ sin(x)+sin(3x)\implies 2sin\left( \cfrac{x+3x}{2} \right)cos\left( \cfrac{x-3x}{2} \right) \\\\\\ 2sin\left( \cfrac{4x}{2} \right)cos\left( \cfrac{-2x}{2} \right)\implies 2sin(2x)cos(-x)\implies 2\boxed{sin(2x)} cos(x) \\\\\\ 2\boxed{2sin(x)cos(x)} cos(x)\implies \underline{4sin(x)cos^2(x)}

User Sonrobby
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