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How do you express sin 3x as a polynom with sin x using the De Moivre formula?
Wiki User
∙ 11y ago
According to de Moivre's formula,
cos3x + isin3x = (cosx + isinx)3 = cos3x + 3cos2x*isinx + 3cosx*i2sin2x + i3sin3x
Comparing the imaginary parts,
isin3x = 3cos2x*isinx + i3sin3x
so that sin3x = 3cos2x*sinx - sin3x
= 3*(1-sin2x)sinx - sin3x
= 3sinx - 4sin3x
Wiki User
∙ 11y ago
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