Question

2（Sinx的4次方）+2(cosx的四次方)+cos2x的平方-3

Answer 1

2(sinx)^4+2(cosx)^4+(cos2x)^2-3

=2[(sinx)^2+(cosx)^2]^2-2*2sin^2xcosx^2+(cos2x)^2-3

=2-(sin2x)^2+(cos2x)^2-3

=(cos2x)^2-(sin2x)^2-1

=cos4x-1