How do you solve cos4u = cos^2 2u-sin^2 2u, within the interval [0,2pi)?

1 Answer
Cesareo R.
Sep 23, 2016

It is an identity so it is true for all u

Explanation:

cos(4u) = cos^2 (2u)-sin^2 (2u)

Using de Moivre's identity

e^(ix) = cosx+isinx

cos^2 (2u)-sin^2 (2u) = ((e^(i2u)+e^(-i2u))/2)^2- ((e^(i2u)-e^(-i2u))/(2i))^2
=(e^(i4u)+2+e^(-i4u))/4+(e^(i4u)-2+e^(-i4u))/4 = (e^(i4u)+e^(-i4u))/2 = cos(4u)

so this is an identity. It is true for all u

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