若x,y为锐角,cosx=3/5,cos(x+y)=根号3/3,求siny的值
问题描述：
若x,y为锐角,cosx=3/5,cos(x+y)=根号3/3,求siny的值
答
x,y为锐角,cosx=3/5,cos(x+y)=√3/3
则由平方关系可得：sinx=4/5,sin(x+y)=√6/3
则：siny=sin[(x+y)-x]
=sin(x+y)cosx-cos(x+y)sinx
=(√6/3)(3/5)-(√3/3)(4/5)
=(3√6-4√3)/15则由平方关系可得：sinx=4/5，sin(x+y)=√6/3tangram_guid_1361715994171x,y为锐角,cosx=3/5,cos(x+y)=√3/3sin²x+cos²x=1，得：sin²x=1-cos²x=16/25，所以，sinx=4/5；sin²(x+y)+cos²(x+y)=1，得：sin²(x+y)=1-cos²(x+y)=2/3，所以，sin(x+y)=√6/3