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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