Question

高中三角函数化简问题求化简函数f(x)=(sin4x+cos4x+sin2xcos2x)/(2-2sinxcosx)si？

Answer 1

f(x)=(sin4x+cos4x+sin2xcos2x)/(2-2sinxcosx)=[(sin2x+cos2x)^2-2sin2xcos2x+sin2xcos2x]/(2-2sinxcosx)=(1+sinxcosx)/2,5,=（sin4x+cos4x+2sin2xcos2x-sin2xcos2x）/(2-sin2x)
=((sin2x+cos2x)^2-sin2xcos2x）/(2-sin2x)
=(1-(1/4)(sin(2x))^2）/(2-sin(2x))
=(1-(sin(2x)/2)^2）/(2-sin(2x))
=(1+sin(2x))/4,2,=((sin2x+cos2x)^2-sin2xcos2x)*2*(1-sinxcosx)
=2(1-sin2xcos2x)(1-sinxcosx)
=2(1-sinxcosx)^2(1+sinxcosx)
其中使用与你相同替代,1,f(x) = {[(sin x)^2+(cos x)^2]^2 - (sin x)^2 * (cos x)^2}/
2(1-sinx * cos x)
= [1-(sin x)^2 * (cos x)^2]/2(1-sinx * cos x)
= (1-sinx * cos x)(1+sinx * cos x)/2(1-sinx * cos x)
=(1+sinx * cos x)/2
=1/2 + sin (2x)/4,0,高中三角函数化简问题
求化简函数f(x)=(sin4x+cos4x+sin2xcos2x)/(2-2sinxcosx)
sin4x cos4x 为 (sinx)^4 (cosx)^4
sin2x 为(sinx)^2 cos2x为 (cosx)^2