lim(cosx-cos4x)/x^4

lim(x趋于派/4)1+sin2x/1-cos4x=lim(x趋于派/4)1+sin2x/1-cos4x=lim(x趋于派/4)1+sin2x/1-cos4x=x→π/4时,直接代入,得sin2x→

lim(x→0)(1-cos4x)/xsinxlim(x→0)(1-cos4x)/xsinxlim(x→0)(1-cos4x)/xsinx点击图片就可以看清楚,加油!1-cos4x=(cos2x)^2

高一二倍角三角函数1.求证：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:（1+sin4x-cos4x)/(1+sin4x+cos4x)-(1+sin4x+c

求极限lim(x→π/4)1+sin2x/1-cos4x求极限lim(x→π/4)1+sin2x/1-cos4x求极限lim(x→π/4)1+sin2x/1-cos4xx→π/4时,直接代入,得sin

cosx+cos2x+cos3x+cos4x=0x=cosx+cos2x+cos3x+cos4x=0x=cosx+cos2x+cos3x+cos4x=0x=请点击图片查看解法

求f(x)=cos4x/(sinx+cosx)^2的最大值求f(x)=cos4x/(sinx+cosx)^2的最大值求f(x)=cos4x/(sinx+cosx)^2的最大值f(x)=cos4x/(s

三角函数二倍角公式的运用?F(X)=COS4x-SIN4x这里的4是COSx.SINx的4次方三角函数二倍角公式的运用?F(X)=COS4x-SIN4x这里的4是COSx.SINx的4次方三角函数二倍

cos4x+4cos2x+3=8(cosx)^4如何证明cos4x+4cos2x+3=8(cosx)^4如何证明cos4x+4cos2x+3=8(cosx)^4如何证明左边=2(cos2x)^2-1+

证明cos4x+4cos2x+3=8cosx^4证明cos4x+4cos2x+3=8cosx^4证明cos4x+4cos2x+3=8cosx^4cos4x+4cos2x+3=2cos^2(2x)-1+

lim(x→0)(cosx)^[4/(x^2)]lim(x→0)(cosx)^[4/(x^2)]lim(x→0)(cosx)^[4/(x^2)]令原式=y则lny=4ln(cosx)/x^2x→0,l

lim(x→0)(cosx)^4/x^2怎么解lim(x→0)(cosx)^4/x^2怎么解lim(x→0)(cosx)^4/x^2怎么解

高中三角函数化简问题求化简函数f(x)=(sin4x+cos4x+sin2xcos2x)/(2-2sinxcosx)sin4xcos4x为(sinx)^4(cosx)^4sin2x为(sinx)^2c

求证(sin4x)/(1+cos4x)*(cos2x)/(1+cos2x)*(cosx)/(1+cosx)=tan(x/2)求证(sin4x)/(1+cos4x)*(cos2x)/(1+cos2x)*

lim（cosx-sinx）/cos2xx→π/4lim（cosx-sinx）/cos2xx→π/4lim（cosx-sinx）/cos2xx→π/4（cosx-sinx）/cos2x=（cosx-s

求lim(x-0)1-cos4x/2sin^2x+xtan^2x的极限求lim(x-0)1-cos4x/2sin^2x+xtan^2x的极限求lim(x-0)1-cos4x/2sin^2x+xtan^

lim1-cos4x/2sin^2x+xtan^2xx趋近于0求极限lim1-cos4x/2sin^2x+xtan^2xx趋近于0求极限lim1-cos4x/2sin^2x+xtan^2xx趋近于0求

lim((x+sinx)/(x-cosx))x->无穷lim((x+sinx)/(x-cosx))x->无穷lim((x+sinx)/(x-cosx))x->无穷lim((x+sinx)/(x-cos

lim(x-cosx)/xx→-∞lim(x-cosx)/xx→-∞lim(x-cosx)/xx→-∞x→-∞lim(x-cosx)/x=lim1-cosx/x=lim1-limcosx/x=1-li

求lim（x-x*cosx)/(x-sinx)求lim（x-x*cosx)/(x-sinx)求lim（x-x*cosx)/(x-sinx)x－xcosx＝x(1－cosx),1－cosx与x^2/2等

lim(tanx-sinx)/x(x→0)的极限；lim(1-cos4x)/xsinx(x→0)的极限lim(tanx-sinx)/x(x→0)的极限；lim(1-cos4x)/xsinx(x→0)的