Question

如何证明 sinα－sinβ=2cos((α+β)/2) ·sin ((α－β)/2)公式

Answer 1

设x=(α+β)/2,y=(α－β)/2,则α=x+y,β =x-y,
sinα－sinβ=sin(x+y)－sin(x-y)=sinxcosy+cosxsiny-(sinxcosy - cosxsiny )
=2cosxsiny =2cos((α+β)/2) ·sin ((α－β)/2)

Answer 2

sinα-sinβ=sin[(α+β)/2+(α－β)/2]-sin[(α+β)/2-(α－β)/2]
=sin[(α+β)/2]cos[(α－β)/2]+cos[(α+β)/2]sin[(α－β)/2]-sin[(α+β)/2]+cos[(α－β)/2]
=2cos[(α+β)/2] ·sin[(α－β)/2]