Question

数学三角函数的题

写一下过程哈
若锐角a，b满足cosa=4/5，cos（a+b）=3/5，则sinb的值是（）
A.17/25 B.3/5 C.7/25 D.1/5

Answer 1

cosa=4/5,sina=3/5
cos(a+b)=cosacosb-sinasinb
3/5=4/5cosb-3/5sinb
3=4cosb-3sinb
(3+3sinb)^2=16cos^2b=16(1-sin^2b)
9+18sinb+9sin^b=16-16sin^2b
25sin^2b+18sinb-7=0
(25sinb-7)(sinb+1)=0
sinb=7/25 sinb=-1(舍去）

Answer 2

cosa=4/5,sina=3/5
cos(a+b)=cosacosb-sinasinb
3/5=4/5cosb-3/5sinb
3=4cosb-3sinb
(3+3sinb)^2=16cos^2b=16(1-sin^2b)
9+18sinb+9sin^b=16-16sin^2b
25sin^2b+18sinb-7=0
(25sinb-7)(sinb+1)=0
sinb=7/25 sinb=-1(舍去）
望采纳