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230问答网 > 已知A,B均为锐角,且cosA=1⼀7,cos(A+B)=-11⼀14,求sinB的值

已知A,B均为锐角,且cosA=1⼀7,cos(A+B)=-11⼀14,求sinB的值

要有过程

2025-03-16 19:20:25

推荐回答（2个）

回答1：

锐角，sinA>0
0sin(A+B)>0

sin²A+cos²A=1
cosA=1/7
所以sinA=4√3/7

同理cos(A+B)=-11/14
sin(A+B)=5√3/14

sinB=[sin(A+B)-A]
=sin(A+B)cosA-cos(A+B)sinA
代入自己算一下

回答2：

cosA=1/7
->
sinA=4√3/7
cos(A+B)=cosAcosB-sinAsinB=-11/14
->
1/7cosB-4√3/7sinB=-11/14
(1)如果是小题,直接目测为cosB=1/2,sinB=√3/2
->
B=π/3
(2)如果是大题,cosB-4√3sinB=-11/2
->
cosB=4√3sinB-11/2
sinB^2+cosB^2=1
->
带入解,可得sinB=-√3/2或√3/2
又因为B为锐角，所以sinB=√3/2
->
cosB=4√3sinB-11/2=1/2
->
B=π/3

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