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Question #34291

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Dec 27, 2016

#sin^4x+cos^4x#

#=(sin^2x+cos^2x)^2-2sin^2xcos^2x#

#=1-2sin^2xcos^2x#

#=1-1/2xx4sin^2xcos^2x#

#=1-1/2xx(2sinxcosx)^2#

#=1-1/2xx((sinx+cosx)^2-(sin^2x+cos^2))^2#

#=1-1/2xx(a^2-1)^2#

#=(2-(a^2-1)^2)/2#

#=(2-a^4+2a^2-1)/2#

#=(1-a^4+2a^2)/2#

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