Question

若cosA=4/5,cos(A+B)=3/5 A,B为锐角，求sinB

Answer 1

cosA=4/5
sinA=3/5=cos(A+B)
因A,B为锐角
A=90-（A+B）
B=90-2A

所以 sinB=cos2A=2（cosA）^2-1=7/25

Answer 2

SinB=sin(A+B-A)
=sin(A+B)cosA-cos(A+B)sinA
cosA=4/5,cos(A+B)=3/5 A,B为锐角
则sinA=[1-(cosA)^2]^(1/2)=3/5
A,B都是锐角，A+B<180，所以sin(A+B)<0
sin(A+B)==[1-(cosA+B)^2]^(1/2)=4/5
SinB=sin(A+B-A)
=sin(A+B)cosA-cos(A+B)sinA
=4/5*4/5-3/5*3/5
=7/25

Answer 3

cos(A+B)=3/5,A,B为锐角
cosA=4/5,sinA=3/5
cosAcosB-sinAsinB=3/5
4√(1-sin^2B)-3sinB=3
16(1-sin^2B)=9(1+sinB)^2，B为锐角
sinB=7/25

Answer 4

∵A,B为锐角∴0<A+B<180°,sin(A+B)=√(1-(cos(A+B)))^2)=4/5
sinA=√(1-(cosA)^2)=3/5
sinB=sin（(A+B)-A) =sin(A+B)cosA-cos(A+B)sinA =(4/5)×(4/5)-(3/5)×(3/5) =7/25