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)^4sin2x 为(sinx)^2 cos2x为 (cosx)^2
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![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)^4sin2x 为(sinx)^2 cos2x为 (cosx)^2](/uploads/image/z/1801701-45-1.jpg?t=%E9%AB%98%E4%B8%AD%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%8C%96%E7%AE%80%E9%97%AE%E9%A2%98%E6%B1%82%E5%8C%96%E7%AE%80%E5%87%BD%E6%95%B0f%28x%29%3D%28sin4x%2Bcos4x%2Bsin2xcos2x%29%2F%282-2sinxcosx%29sin4x+cos4x+%E4%B8%BA+%28sinx%29%5E4++%28cosx%29%5E4sin2x+%E4%B8%BA%28sinx%29%5E2++cos2x%E4%B8%BA+%28cosx%29%5E2)
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)^4sin2x 为(sinx)^2 cos2x为 (cosx)^2
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
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)^4sin2x 为(sinx)^2 cos2x为 (cosx)^2
f(x)=(sin4x+cos4x+sin2xcos2x)/(2-2sinxcosx)=[(sin2x+cos2x)^2-2sin2xcos2x+sin2xcos2x]/(2-2sinxcosx)=(1+sinxcosx)/2
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
=(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
=((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