<s>[INST] Application Of De Moivre & Euler's Formulae Can anyone show me how I can prove that $\sin x \cos (3x) = \frac{1}{4} \sin (7x) - \frac{1}{4}\sin (5x) + \frac{1}{2}\sin x$?
I tried using Euler's formulae
$$\sin x= \frac{e^{ix}-e^{-ix}}{2i}$$ 
and
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
but the simplification didn't help at all.
PS: Simplify starting from the left. 
 [/INST] It should be $\sin { x } \cos ^{ 2 }{ 3x } $
$$\frac { 1 }{ 4 } \sin { 7x } -\frac { 1 }{ 4 } \sin { 5x } +\frac { 1 }{ 2 } \sin { x } =\frac { 1 }{ 4 } \left( \sin { 7x } -\sin { 5x } +2\sin { x }  \right) =\\ =\frac { 1 }{ 4 } \left( 2\sin { \frac { 7x-5x }{ 2 } \cos { \frac { 7x+5x }{ 2 }  }  } +2\sin { x }  \right) =\frac { 1 }{ 2 } \left( \sin { x } \left( \cos { 6x } +1 \right)  \right) =\\ =\frac { \sin { x }  }{ 2 } \left( 2\cos ^{ 2 }{ 3x }  \right) =\sin { x } \cos ^{ 2 }{ 3x } \\ $$
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