cos2x cosπ/5+sin2x sinπ/5化简最详细的步骤?
cos2x cosπ/5+sin2x sinπ/5化简最详细的步骤?
化简:2cos2x+2sin^2 x+cos(-x)分之sin2x+sin(π-x)=___________
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
已知sin(π-α)=4/5 ,α属于(0.2).求函数f(x)=5/6 cosαsin2x-½cos2x
求证 sinx(cos^2 2x-sin^2 2x) + 2cosx cos2x sin2x= sin 5x
两个个数学证明题1.证明:cos2x+sin2x=√2sin(sin2x+π/4)2.证明:αsinαx+cosαx=√
要得到函数y=2cos(x+π/6)sin(π/3-x)-1的图像只需将函数 y=1/2sin2x+√3/2cos2x
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
已知tan(x-π/4),求(sin2x+2cos2x)/(2cos²x-3sin2x-1)的值
已知cos(π/4+x)=3\5,求(sin2x-2sin^x)/1-tanx的值
已知cos(π/4-x)=-3/5,求sin2x/sin(x+π/4)