化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x)

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：2024/11/17 14:04:10

已知sin^2x+cos^2x=1 分子配方 1-sin^4x-cos^4x=1+2sin^2xcos^2x-(sin^2x+cos^2x)^2=1+2sin^2xcos^2x-1=2sin^2xcos^2x 分母先用立方和公式,再配方 1-sin^6-cos^6x=1-(sin^2x+cos^2x)*(sin^4x-sin^2xcos^2x+cos^4x)=1-(sin^4x-sin^2xcos^2x+cos^4x)=1+3sin^2xcos^2x-(sin^2x+cos^2x)^2=3sin^2xcos^2x 因此原式=2/3

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 化简（1)√3sin x+cos x (2）√2(sin x-cos x) （3）√2cos x-√6sin x 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2 sin(1/x)-cos(1/x)/x (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1 化简 1/2cos x-根号3/2sin x 根3sin x+cos x 2cos x (sin x -cos x)+1 f x =sin(pai*x/4-pai/6)-2(cos pai*x/8)^2+1 三角函数证明及化简题1.证明sin^6x＋cos^6x＝（sin^2x＋cos^2x)（cos^4x－cos^2xsin