已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B 已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb
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![已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B 已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb](/uploads/image/z/5255373-21-3.jpg?t=%E5%B7%B2%E7%9F%A52tanA%3D3tanB%2C%E6%B1%82%E8%AF%81tan%EF%BC%88A%2BB%EF%BC%89%3Dsin2B%2F5-cos2B+%E5%B7%B2%E7%9F%A5a+%2Cb%E4%B8%BA%E9%94%90%E8%A7%92%2Ccosa%3D4%2F5%2Ctan%EF%BC%88a-b%EF%BC%89%3D1%2F3+%2C%E6%B1%82cosb)
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已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B 已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb
已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B
已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb
已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B 已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb
首先,第一题的答案是错误的,tan(A+B)=(tanA+tanB)/(1-tanA*tanB),把tanA=3tanB/2代入,得tan(A+B)=5tanB/(2-3tanB*tanB),又因为tanB=sinB/cosB,易得tan(A+B)=5sinBcosB/(2cosB*cosB-3sinB*sinB),又因为cosB*cosB=(cos2B+1),sinB*sinB=(1-cos2B),式子最终得5sin2B/(5cos2B-1)
第二题,因为a,b是锐角,cosa=4/5,所以tana=3/4,tanb=tan[a-(a-b)]=[tana-tan(a-b)]/[1+tana*tan(a-b)]=1/3,所以cosb=3(根号10)/10
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