sin^6 a+cos^6 a+3sin^2 acos^2a=(sin^2 a+cos^2 a)(sin^4 a+cos

已知a(-0,兀且2SIN A-SIN ACOS A-3COS A=0求SIN(+兀/4)/SIN 2A+COS 2A+ 化简sin^6a+cos^6+3sin^2acos^2a的结果是? 求证sin^4a+cos^4a=1-2sin²acos²a 已知(4sin a－2cos a)／(5cos a +3sin a)=6/11,求cos＾4 a－sin＾4 a 化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)== (1-sin^6a-cos^6a)/(sin^2a-sin^4a) 证明Cos^A-Sin^A=1-2Sin^A=2Cos^A-1=cos^a-sin^a 已知 sin a-cos a=1/2,则 sin^3 a-cos^3 a (cos a)^4+(sin a)^2*(sin a)^2+(sin a)^2=? 已知tan a =-4,求下列各式的值 (1)sin^2 a (2)3sin acos a (3)cos^2 a-sin 急.三角函数：化简（sin^2a*cos^4a+sin^4acos^2a)/(1-sin^4a-cos^4a) 求证[1]1-2sin acos a/cos²a -sin²a =1-tan a/1+tan a.[