作业帮求证sin²x-sin²y=sin(x+y)sin(x-y)
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求证sin²x-sin²y=sin(x+y)sin(x-y)
答：
sin²x-sin²y
=(sinx+siny)(sinx-siny)
=2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2]
=sin(x+y)*sin(x-y)
所以：sin²x-sin²y=sin(x+y)*sin(x-y)
再问： =2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2] =sin(x+y)*sin(x-y) 这两部为什么直接相等了？
再答： 因为两倍角公式：sin2A=2sinAcosA 所以： =2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2] =2sin[(x+y)/2]cos[(x+y)/2] * 2cos[(x-y)/2]sin[(x-y)/2] 把两个cos的因子调换一下位置 =sin(x+y)*sin(x-y)
sin²x-sin²y
=(sinx+siny)(sinx-siny)
=2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2]
=sin(x+y)*sin(x-y)
所以：sin²x-sin²y=sin(x+y)*sin(x-y)
再问： =2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2] =sin(x+y)*sin(x-y) 这两部为什么直接相等了？
再答： 因为两倍角公式：sin2A=2sinAcosA 所以： =2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2] =2sin[(x+y)/2]cos[(x+y)/2] * 2cos[(x-y)/2]sin[(x-y)/2] 把两个cos的因子调换一下位置 =sin(x+y)*sin(x-y)
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