已知a B为锐角,且cosa=4/5,cos(a+B)=-16/65,求cosB的值,急用

题目

已知a B为锐角,且cosa=4/5,cos(a+B)=-16/65,求cosB的值,急用

答案

cosa =4/5
a 为锐角,所以sina>0,
sina=√(1- cos²a) = 3/5
cos(a+b)= -16/65,
a、b为锐角,所以 a+b < 180°,sin(a+b) >0
即 sin(a+b)= √【1 - cos²（a+b)】 = 63/65
cos(a+b)= cosacosb - sinasinb = (4/5)cosb - (3/5)sinb =-16/65 (1)
sin(a+b)= sinacosb +cosasinb = (3/5)cosb + (4/5)sinb = 63/65 (2)
(1)× 4 + (2)× 3得
5cosb = 125/65
cosb =5/13