急.三角函数:化简(sin^2a*cos^4a+sin^4acos^2a)/(1-sin^4a-cos^4a)
急.三角函数:化简(sin^2a*cos^4a+sin^4acos^2a)/(1-sin^4a-cos^4a)
求证sin^4a+cos^4a=1-2sin²acos²a
已知a(-0,兀且2SIN A-SIN ACOS A-3COS A=0求SIN(+兀/4)/SIN 2A+COS 2A+
sin²a+cosˇ4a+sin²acos²a 化简
化简cos^2 a(2cos^2+3)-sin^2 a(2cos^2+3)-4sin^2 acos^2 a+3
已知cos2a=(1/4),求cos^4a+sin^4+sin^2acos^2a的值
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
(1-sin^6a-cos^6a)/(sin^2a-sin^4a)
化简 1-sin^4a-cos^4a/cos^2a-cos^4a
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
(cos a)^4+(sin a)^2*(sin a)^2+(sin a)^2=?