已知A,B为锐角,sinA=根号5⼀5,tanB=3,求cos(A-B)和A-B

2025-06-29 02:09:45
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回答1:

A,B为锐角sinA=√5/5cosA=√[1-(sinA)^2=2√5/5tanB=3sinB/cosB=3(sinB)^2+(cosB)^2=1所以sinB=3/√10cosB=1/√10cos(A-B)=cosAcosB+sinAsinB=√2/2cos(A+B)=cosAcosB-sinAsinB=-√2/10