Prove that $cos(4x)=8 cos(x)^4 - 8 cos(x)^2 +1$ ?

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Answer 1 · 2016-10-11T18:53:11

$cos(4x)=8 cos(x)^4 - 8 cos(x)^2 +1$

Using de Moivre's identity

$e^(ix)=cos(x)+isin(x)$ we obtain

$e^(4ix) = cos(4x)+isin(4x) = (cos(x)+isin(x))^4$

but

$(cos(x)+isin(x))^4=Cos(x)^4 - 6 Cos(x)^2 Sin(x)^2 + Sin(x)^4 + i (4 Cos(x)^3 Sin(x) - 4 Cos(x) Sin(x)^3)$

Keeping the real component

$cos(4x)=Cos(x)^4 - 6 Cos(x)^2 Sin(x)^2 + Sin(x)^4 =$
$=Cos(x)^4+ 6 Cos(x)^2(1-cos(x)^2)+(1-cos(x)^2)^2=$
$=8 cos(x)^4 - 8 cos(x)^2 +1$

Prove that cos(4x)=8 cos(x)^4 - 8 cos(x)^2 +1cos(4x)=8 cos(x)^4 - 8 cos(x)^2 +1 ?