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Question Number 83654 by jagoll last updated on 05/Mar/20
solve this equation sin^2 x−sin^4 x=cos^2 x−cos^4 x
solvethisequationsin2x−sin4x=cos2x−cos4x
Commented by jagoll last updated on 05/Mar/20
⇔ cos^4 x−sin^4 x = cos^2 x−sin^2 x ⇒ cos^2 x−sin^2 x= cos^2 x−sin^2 x ∴ always true for x ∈R
⇔cos4x−sin4x=cos2x−sin2x⇒cos2x−sin2x=cos2x−sin2x∴alwaystrueforx∈R
Answered by MJS last updated on 05/Mar/20
cos^2 x =1−sin^2 x sin^2 x −sin^4 x =1−sin^2 x −(1−sin^2 x)^2 sin^2 x −sin^4 x =(1−sin^2 x)sin^2 x lhs=rhs true ∀x∈R
cos2x=1−sin2xsin2x−sin4x=1−sin2x−(1−sin2x)2sin2x−sin4x=(1−sin2x)sin2xlhs=rhstrue∀x∈R
Commented by jagoll last updated on 05/Mar/20
yess...
yess…
Answered by $@ty@m123 last updated on 05/Mar/20
sin^2 x(1−sin^2 x)=cos^2 x(1−cos^2 x) sin^2 xcos^2 x=cos^2 xsin^2 x
sin2x(1−sin2x)=cos2x(1−cos2x)sin2xcos2x=cos2xsin2x

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