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.
Tuesday 7 August 2018
trigonometry - Application Of De Moivre & Euler's Formulae
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