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

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

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Answer 1 · 2016-09-23T10:32:09

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

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