2(sinx)^4+3/4sin(2x)^2+5cosx^4-1/2cos4x-1/2cos2x=2(1+cosx^2)
7个回答

求证题算法

3/4sin(2x)^2=3/4(2sinxcosx)^2=3(sinxcosx)^2

所以左边=2(sinx)^4+4(sinxcosx)^2+2(cosx)^4-(sinxcosx)^2+3(cosx)^4-1/2[2(cos2x)^2-1]-1/2cos2x

=2[(sinx)^2+(cosx)^2]^2-(sinxcosx)^2+3(cosx)^4-(cos2x)^2+1/2-1/2[2(cosx)^2-1]

=2*1^2-(sinxcosx)^2+3(cosx)^4-(cos2x)^2+1/2-(cosx)^2+1/2

=3-(sinxcosx)^2+3(cosx)^4-[2(cosx)^2-1]^2-(cosx)^2

=3-(sinxcosx)^2+3(cosx)^4-4(cosx)^4+4(cosx)^2-1-(cosx)^2

=2-(sinx)^2(cosx)^2-(cosx)^4+(cosx)^2+2(cosx)^2

=2+(cosx)^2[1-(sinx)^2]-(cosx)^4+2(cosx)^2

=2+(cosx)^4-(cosx)^4+2(cosx)^2

=2+2(cosx)^2

=2[1+(cosx)^2]

=右边

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