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sin(nπ/2)*sin(nπ/3)+sin(nπ/2)*sin(nπ/6)为什么可以化解成cos(nπ/6)+cos(nπ/3)
题目详情
sin(nπ/2)*sin(nπ/3) + sin(nπ/2)*sin(nπ/6) 为什么可以化解成cos(nπ/6)+cos(nπ/3)
▼优质解答
答案和解析
sin(nπ/2)sin(nπ/3)+sin(nπ/2)*sin(nπ/6)
=(1/2)[cos(nπ/2-nπ/3)-cos(nπ/2+nπ/3)] +(1/2)[cos(nπ/2-nπ/6)-cos(nπ/2+nπ/6)]
=(1/2)[cos(nπ/6)-cos(5nπ/6)]+(1/2)[cos(nπ/3)-cos(2nπ/3)]
cos(5nπ/6)=cos(nπ-nπ/6),cos(2nπ/3)=cos(nπ-nπ/3)
n为奇数时
cos(5nπ/6)=cos(nπ-nπ/6)=cos(π-nπ/6)=-cos(nπ/6)
cos(2nπ/3)=cos(nπ-nπ/3)=cos(π-nπ/3)=-cos(nπ/3)
原式= cos(nπ/6) +cos(nπ/3)
n为偶数时
cos(5nπ/6)=cos(nπ-nπ/6)=cos(-nπ/6)=cos(nπ/6)
cos(2nπ/3)=cos(nπ-nπ/3)=cos(-nπ/3)=cos(nπ/3)
原式=0
积化和差公式
sinxsiny=(1/2)[cos(x-y)-cos(x+y)]
cos(x-y)=cosxcosy+sinxsiny
cos(x+y)=cosxcosy-sinxsiny
=(1/2)[cos(nπ/2-nπ/3)-cos(nπ/2+nπ/3)] +(1/2)[cos(nπ/2-nπ/6)-cos(nπ/2+nπ/6)]
=(1/2)[cos(nπ/6)-cos(5nπ/6)]+(1/2)[cos(nπ/3)-cos(2nπ/3)]
cos(5nπ/6)=cos(nπ-nπ/6),cos(2nπ/3)=cos(nπ-nπ/3)
n为奇数时
cos(5nπ/6)=cos(nπ-nπ/6)=cos(π-nπ/6)=-cos(nπ/6)
cos(2nπ/3)=cos(nπ-nπ/3)=cos(π-nπ/3)=-cos(nπ/3)
原式= cos(nπ/6) +cos(nπ/3)
n为偶数时
cos(5nπ/6)=cos(nπ-nπ/6)=cos(-nπ/6)=cos(nπ/6)
cos(2nπ/3)=cos(nπ-nπ/3)=cos(-nπ/3)=cos(nπ/3)
原式=0
积化和差公式
sinxsiny=(1/2)[cos(x-y)-cos(x+y)]
cos(x-y)=cosxcosy+sinxsiny
cos(x+y)=cosxcosy-sinxsiny
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