sin x *cos y-cos x*siny 用Mathematica怎么化简
sin x *cos y-cos x*siny 用Mathematica怎么化简
求证:sin(x-y)/(sinx-siny)=cos[(x-y)/2]/cos[(x+y)/2]
若cos(x+y)cosy+sin(x+y)siny=0,则cosx=
证明 [sin(2x+y)/sinx]-2cos(x+y)=siny/sinx
请问,如何证明sinx+siny=2*sin(x+y/2)*cos(x-y/2)
已知cos(x+y)cosy+sin(x+y)siny=4/5,求tanx的值
求证sinx+siny=2sin(x+y)/2*cos(x-y)/2
求证|sinx-siny|=|2sin[(x-y)/2]cos[(x+y)/2]|
求证:sin(2x+y)/sinx-2cos(x+y)=siny/sinx
为什么sin(x+y) - sinx = 2cos(x+y/2)siny/2
已知x-y=派/3,且sin x-siny=1/2,求cos(x+y).
已知sinx+siny=1/3,cosx-cosy=1/5,求cos(x+y),sin(x-y).