"己知cosA=1/7,cos(A B)=-(11/14)且A,(0,90),求cosB的值"

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"己知cosA=1/7,cos(AB)=-(11/14)且A,(0,90),求cosB的值""己知cosA=1/7,cos(AB)=-(11/14)且A,(0,90),求cosB的值""己知cosA=

"己知cosA=1/7,cos(A B)=-(11/14)且A,(0,90),求cosB的值"
"己知cosA=1/7,cos(A B)=-(11/14)且A,(0,90),求cosB的值"

"己知cosA=1/7,cos(A B)=-(11/14)且A,(0,90),求cosB的值"
因为 cosA=1/7,所以 sinA = (4√3)/7,
cosC=cos[π-(A+B)]=-cos(A+B)=11/14,所以 sinC= (5√3)/14.cosB=cos[π-(A+C)]=-cos(A+C)=sinAsinC-cosAcosC =[(4√3)/7]*[(5√3)/14]-(1/7)*(11/14)=1/2.