两角和差公式sinx-siny=2sin[(x-y)/2]cos[(x+y)/2]推导.

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两角和差公式
sinx-siny
=2sin[(x-y)/2]cos[(x+y)/2]
推导.

sinx-siny
=sin[(x-y)/2 + (x+y)/2] -sin[(x+y)/2 -(x-y)/2] （变形后再利用两角和与差的正弦公式）
=sin[(x-y)/2]cos[(x+y)/2] +cos[(x-y)/2]sin[(x+y)/2] -{sin[(x+y)/2]cos[(x-y)/2] -cos[(x+y)/2]sin[(x-y)/2]}
=2sin[(x-y)/2]cos[(x+y)/2]

两角和差公式sinx-siny=2sin[(x-y)/2]cos[(x+y)/2]推导. 证明 [sin(2x+y)/sinx]-2cos(x+y)=siny/sinx 求证：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) 求证sinx+siny=2sin(x+y)/2*cos(x-y)/2 求证|sinx-siny|=|2sin[(x-y)/2]cos[(x+y)/2]| 为什么sin（x+y) - sinx = 2cos(x+y/2)siny/2 求证：sin（x-y）/（sinx-siny）=cos[（x-y）/2]/cos[(x+y)/2] 证明sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((x+y)/2)sin((x+ 证明sin(x+y)sin(x-y)=（sinx）^2-(siny）^2. 证明cosx(cosx-cosy)+sinx(sinx-siny)=2sin(x-y)/2 证明sin(x+y)sin(x-y)=sinx-siny