设t=sinx+cosx=2sin(x+π4),∵x∈(0,π3),
设t=sinx+cosx=2sin(x+π4),∵x∈(0,π3),
设f(x)=2cosx.sin(x+π/3)-根号3 sin平方x+sinx.cosx
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx)
已知函数f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
f(x)=cos(2x-π/3)+2sin(x-π/4)怎么化简成(sinx-cosx)(sinx+cosx)
已知sinx+cosx/2sinx-3cosx=3 求 2sin²x-3sin(3π+x)cos(π-x)-3
求定积分(0到π/2)sin^3x/(sinx+cosx)dx=?
利用cosx=sin(π/2-x),sin'x=cosx,证明(cosx)'=-sinx
证明定积分(0到π/2)sin^3x/(sinx+cosx)dx=定积分(0到π/2)cos^3x/(sinx+cosx
已知f(x)=2cosx*sin(x+π/6)+√3sinx*cosx-sin^2x.设三角形ABC的内角A满足f(2)
已知函数y=(cos(x-(π/4))-0.5)/(1+sinx+cosx),0≤x≤π/2,设t=sinx+cosx