求证:cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)
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:2024/10/09 19:23:22
求证:cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)
还有,csc^4a(1-cos^4a)-2cot^2a
还有,csc^4a(1-cos^4a)-2cot^2a
cos^8x-sin^8x-cos2x
=(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x
=(cos^4x+sin^4x)*1*cos2x-cos2x
=[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2-1]cos2x
=-(2sinxcosx)^2/2*cos2x
=-sin^2 2x/2*co2x
=-(1+cos4x)*cos2x/4
1/8(cos6x - cos2x)
=1/8*(-2)*sin(8x/2)sin(4x/2)
=-1/4sin4xsin2x
=-1/4*(2sin2xcos2x)sin2x
=-1/4*2(sin2x)^2cos2x
=-1/4(1+cos4x)cos2x
所以cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)
=(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x
=(cos^4x+sin^4x)*1*cos2x-cos2x
=[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2-1]cos2x
=-(2sinxcosx)^2/2*cos2x
=-sin^2 2x/2*co2x
=-(1+cos4x)*cos2x/4
1/8(cos6x - cos2x)
=1/8*(-2)*sin(8x/2)sin(4x/2)
=-1/4sin4xsin2x
=-1/4*(2sin2xcos2x)sin2x
=-1/4*2(sin2x)^2cos2x
=-1/4(1+cos4x)cos2x
所以cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)
求证 cos^8(x)-sin^8(x)=cos2x【1-1/2sin^2(2x)】
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x
求证 sinx(cos^2 2x-sin^2 2x) + 2cosx cos2x sin2x= sin 5x
f(x)=(1+cos2x)/[4sin(pai/2+x)]-asin(x/2)cos(pai-x/
函数f(x)=(1+cos2x+8sin^2x)/sin2x的值域为
cos^2 x=(cos2x-1)/2
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
已知sin(x/2)+cos(x/2)=1/2,则cos2x=?
求一道高数里的函数题f(sin x)=cos2x+1,求f(cos x).
已知函数f(x)=〔3-2cos^2(x)-8sin^4(x)〕/cos2x.求函数的值域
化简:2cos2x+2sin^2 x+cos(-x)分之sin2x+sin(π-x)=___________