求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值,

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/09/22 19:42:33

(1-cos^6x-sin^6x)/(1-cos^4x-sin^4x)
=[1-(cos^6x+sin^6x)]/[1-(cos^4x-sin^4x)]
=[1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)(cos^2x+sin^2x)]
=[1-(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)]
=[1-(cos^4x+2cos^2xsin^2x+sin^4x-3cos^2xsin^2x)]/[1-cos^2x+sin^2x]
={1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x]}/[sin^2x+sin^2x]
={1-[1-3cos^2xsin^2x]}/(2sin^2x)
=(1-1+3cos^2xsin^2x)/(2sin^2x)
=3cos^2xsin^2x/(2sin^2x)
=3cos^2x/2
再问：看不太懂这么复杂吗
再答：用平方差和立方差

求∫1/(sin^4x+cos^4x)dx, 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 已知sin^3x+cos^3x=1,求sinx+cosx的值与sin^4x+cos^4x的值已知cos 2x=3/5,求sin^4 x+cos^4 x的值求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2 已知sin（x+π/6）的 =1/4,求sin（7π/6 +x）+cos（5π/6 -x）的平方的值已知sin(x+π/6)=1/4,求sin(7π/6+x)+cos(5π/6-x)平方的值已知sin（x+π/6）＝1/4,求sin（7π/6+x）+cos（11π/6-x）的值已知sin(x+π/6)=1/4,求sin(x+7π/6)+cos^2 (11π/6-x)的值 sin(1/x)-cos(1/x)/x 已知tanx=2,(1)求2/3sin^2x+1/4cos^2x的值(2)求2sin^2x-sinxcosx+cos^2