tanα＝4√3,cos(a＋β)＝11／14...a,β为锐角.求cosβ

题目

tanα＝4√3,cos(a＋β)＝11／14...a,β为锐角.求cosβ

答案

a,β为锐角tanα＝4√3=>sina=4√3cosa=>sin^2a=48cos^2a=>sina=4√3/7,cosa=1/7cos(a＋β)＝11／14=>sin(a＋β)=5√3/14cosβ=cos[(a+β)-a]=cos(a＋β)cosa+sin(a＋β)sina=11／14*1/7+5√3/14*4√3/7=71/98...