由和差公式得 sin(A+B)=sinAcosB+cosAsinB=3/5, sin(A-B)=sinAcosB-cosAsinB=1/5, 上面两式相加减得sinAcosB=2/5,cosAsinB=1/5,再将上两式相除得 tanA/tanB=2, 又cos(A+B)=-√(1-(sin(A+B))^2)=-4/5,(由于是锐角三角形C90,cos(A+B)取负值) cos(A-B)=√(1-(sin(A-B))^2)=2√6/5, cos(A+B)=cosAcosB-sinAsinB=-4/5, cos(A-B)=cosAcosB+sinAsinB=2√6/5, 上面两式相加减得c...
|