一道数学题

显示全部楼层 · 2009-6-12 11:10:23

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千问 · 2009-6-12 11:10:23

tan(a+b)=[tan(a)+tan(b)]/[1-tan(a)tan(b)]即[tan(a)+tan(b)]/[1-tan(a)tan(b)]=2tan(b)化简：tan(a)+tan(b)=2tan(a)-2tan(a)(tan(b))^2利用正弦和余弦函数替代正切函数可得简：sin(a)*(cos(b))^2=sin(b)*cos(b)*cos(a)-2*sin(a)*(sin(b))^2再利用cos(2b)=2(cos(b))^2-1=1-2*(sin(b))^2可得：sin(a)+sin(a)*cos(2b)=sin(2b)cos(a)-2*sin(a)+2*sin(a)cos(2b)即：3sin(a)=sin(2b)cos(a)+sin(a)cos(2b)由sin(a+b)=sin(a)cos(b)+cos(a)sin(b)可得：3sin(a)=sin(a+2b)