若x=π/12,则cos^4x-sin^4x=

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若x=π/12,则cos^4x-sin^4x=若x=π/12,则cos^4x-sin^4x=若x=π/12,则cos^4x-sin^4x=cos^4x-sin^4x=(cos^2x+sin^2x)(c

若x=π/12,则cos^4x-sin^4x=
若x=π/12,则cos^4x-sin^4x=

若x=π/12,则cos^4x-sin^4x=
cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)
cos^2x+sin^2x=1
cos^2x-sin^2x= cos (2x)
所以 cos^4x-sin^4x=cos (2x)
若x=π/12
cos^4x-sin^4x=cos (2x)=cos (π/6)=(根号3)/2

cos^4x-sin^4x=(cos^2x-sin^2x)(cos^2x+sin^2x)
=cos2x
x=π/12,cos2x=cosπ/6=√3/2.