cos^4x-sin^4x+2sin^2x=cos^2x-sin^2x+2sin^2x=1
cos^4x-sin^4x+2sin^2x=?
2个回答
相关问题
-
求证(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x
-
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
-
S=(cos^2 x+cos^4 x+.+cos^2n x)+(sin^2x+sin^4 x+.+sin^2n x)数列
-
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
-
sin^4x cos^2x+cos^4x sin^2 x=sin^2x*cot^2x 怎么得来
-
(1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1
-
化简f(x)=(√(sin(x/2))^4+4(cos(x/2))^2))-(√(cos(x/2))^4+4(sin(x
-
证明(sin4x+sin2x)/(cos4x+cos2x)=tan3x
-
cos^4 x-sin^4 x+2 sin^2 x化简
-
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x