cos6x-cos4x

隆侮学3241求使得sin4xsin2x - sinxsin3x=a在[0,π)有唯一解的a. -
寿诗娄15861131390 ______[答案] 令 f(x)=sin4xsin2x-sinxsin3x=− 1 2(cos6x−cos2x)+ 1 2(cos4x−cos2x)= 1 2(cos4x-cos6x), 则有f′(x)=3sin6x-2sin4x,令f′(x)=0,可得x=0 或 x= π 2, 即f′(0)=0,f′( π 2)=0,而且还有f′(π)=0. 由于f′(x)在x=0的左侧小于0,右侧大于0,故f(0)是函数的极...

隆侮学3241求证:cos^8x - sin^8x - cos2x = 1/8(cos6x - cos2x)
寿诗娄15861131390 ______ 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...

隆侮学3241三角函数化简分母是1 - sin4x - cos4x 分子是1 - sin6x - cos6x 4和6都是次方 -
寿诗娄15861131390 ______[答案] 原式=1-sin^(6)x-cos^(6)x/1-sin^(4)x-cos^(4)x=1-{[sin^(2)x]^(3)+[cos^(2)x]^(3)}/1-sin^(4)x-cos^(4)x=1-[sin^(2)x+cos^(2)x]*[sin^(4)x-sin^(2)x*cos^(2)x+cos^(4)x/1-sin^(4)x-cos^(4)x=1-sin^(4)x-cos^(4)x+...

隆侮学3241求证:cos^8x - sin^8x - cos2x = 1/8(cos6x - cos2x)还有,csc^4a(1 - cos^4a) - 2cot^2a -
寿诗娄15861131390 ______[答案] 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... =-(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) =-1/4sin4xsin2...

隆侮学3241cos2x+cos4x+1=cos6x 求x解集 -
寿诗娄15861131390 ______ cos4x+1=2cos²2x 由cos2x+cos4x+1=cos6x得2cos²2x+cos2x=cos4xcos2x-sin4xsin2x2cos²2x+cos2x(1-cos4x)=-sin4xsin2x2cos²2x+2cos2xsin²2x=-sin4xsin2x2cos²2x=-4cos2xsin²2x cos2x(cos2x+2sin²2x)=0 因为 cos2x+2sin²2x≠0 则 cos2x=0时,即 x=kπ+π/4

隆侮学3241求证cos2x+cos4x+cos6x=10 -
寿诗娄15861131390 ______[答案] 怎么可能,有问题的式子怎么证明相等啊,不信我举反例就可推翻了 x=30度时=cos60+cos120+cos180=0.5-0.5-1=-1 x=60度时=cos120+cos240+cos360=-0.5-0.5+1=0

隆侮学3241求证明.关于cos和sin 题目有点长0 0【cos(2x) + cos(4x) + cos(6x)】 / 【sin(2x) + sin(4x) + sin(6x) 】 = cot(4x) -
寿诗娄15861131390 ______[答案] 原式=[(cos2x+cos6x)+cos4x]/[(sin2x+sin6x)+sin4x] =[2cos4xcos2x+cos4x]/[2sin4xcos2x+sin4x] =[cos4x(2cos2x+1)]/[sin4x(2cos2x+1)] =cos4x/sin4x =cot4x.

隆侮学32411 - sin6x - cos6x/1 - sin4x - cos4x -
寿诗娄15861131390 ______ (1-sin6x-cos6x)/(1-sin4x-cos4x) =[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)=(1-1+sin^2xcos^2x+sin^2xcos^2x)/(1-1+2sin^2xcos^2x)=3sin^2xcos^2x/2sin^2xcos^2x=3/2

隆侮学3241(1 - cos^4X - sin^4x)∕(1 - cos^6x - sin^6x)=化简(急) -
寿诗娄15861131390 ______[答案] 1-cos^4x-sin^4x =1-(cos^4x+sin^4x) =1-[(cos^2x+sin^2x)^2-2cos^2xsin^2x] =1-(1-2cos^2xsin^2x) =2cos^2xsin^2x 1-cos^6x-sin^6x =1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x) =1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x] =1-[1-3cos^2xsin^2x] =3cos^2...

隆侮学3241lim(x趋向0)[1 - cos(1 - cos2x)]/x^4怎么解? -
寿诗娄15861131390 ______ lim(x→0)[1-cos(1-cos2x)]/x^4 =lim(x→0)1/2*(1-cos2x)^2/x^4(等价无穷小代换(1-cosx~x^2/2)) =lim(x→0)1/2*1/2*(2x)^2/x^4(等价无穷小代换(1-cosx~x^2/2)) =1