方程X^2sinθ+Xcosθ-π/4=0的两个根为x1.x2,且x1≠x2,则过点（x1^2,x1)和(x2^2,x2)的直线l被圆x^2+y^2=1截得的弦长________求详解

求曲线x^2-6xcosθ-4y+9cos^2θ+8sinθ=0(θ为参数)的焦点轨迹方程 求证sin^4x+cos^4x=1-2sin^2xcos^2x sin^4x+cos^4x=1-2sin^2Xcos^2X谢谢额 化简sin^4x+sin^2xcos^2x+cos^2x=?(请写过程) 已知曲线C方程为16x^2+4y^2-32xcosθ-16y(sinθ)^2-4(sin2θ)^2=0 (θ属于R),求曲线C中心的轨迹的普通方程 积分 sin^4xcos^2x dx利用三角学解以下积分(如图)Hints:sin^4xcos^2x = sin^2x(sinxcosx)^2 sin^4+sin^2xcos^2x+cos^2x 证明 =1 sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简） 若函数f(x)=xcos(2x-π/4)-1/2sin(2x-π/4)(0 求方程[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0的通解, sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos^2x+cos^4x 是根据什么化简的? 求证 cos*xcos*y + sin*xsin*y + sin*xcos*y + xin*ycos*x = 1cos*xcos*y + sin*xsin*y + sin*xcos*y + sin*ycos*x = 1注意：[*] 的意思是 [ ^2 ]写下左右过程.. sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x 那么这个依据呢?老师 求sin^4x+cos^4x=1-2sin^2xcos^2x的详细答案! 求函数y=2sin xcos x+2sin x+2cos x+4的值域 已知tanx=-3/4,求2sin^2 x+3sin xcos x-cos^2 x 设x∈(0,π/2),如何求sin^2xcos^2x+2/sin^2xcos^2x-2的最小值.为什么用基本不等式sin^2xcos^2x+2/sin^2xcos^2x-2≥2√2-2会得到错误的答案呢?求解惑!（答案是25/4) 1/(sin^4 xcos^4 x)积分