1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?2.已知tana*tanb=13/7,tan[(a+b)/2]=(根号6)/2,求cos(a-b)3.已知a,b∈(0,π),且tan(a/2)=1/2,sin(a+b)=5/13,求cosb
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![1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?2.已知tana*tanb=13/7,tan[(a+b)/2]=(根号6)/2,求cos(a-b)3.已知a,b∈(0,π),且tan(a/2)=1/2,sin(a+b)=5/13,求cosb](/uploads/image/z/3975343-7-3.jpg?t=1.%E5%B7%B2%E7%9F%A5tan%28a%2Bb%29%3D3%2Ctan%28a-b%29%3D2%2C%E5%88%99sin4a%3D%3F2.%E5%B7%B2%E7%9F%A5tana%2Atanb%3D13%2F7%2Ctan%5B%28a%2Bb%29%2F2%5D%3D%EF%BC%88%E6%A0%B9%E5%8F%B76%EF%BC%89%2F2%2C%E6%B1%82cos%28a-b%293.%E5%B7%B2%E7%9F%A5a%2Cb%E2%88%88%EF%BC%880%2C%CF%80%EF%BC%89%2C%E4%B8%94tan%28a%2F2%29%3D1%2F2%2Csin%28a%2Bb%29%3D5%2F13%2C%E6%B1%82cosb)
1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?2.已知tana*tanb=13/7,tan[(a+b)/2]=(根号6)/2,求cos(a-b)3.已知a,b∈(0,π),且tan(a/2)=1/2,sin(a+b)=5/13,求cosb
1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?
2.已知tana*tanb=13/7,tan[(a+b)/2]=(根号6)/2,求cos(a-b)
3.已知a,b∈(0,π),且tan(a/2)=1/2,sin(a+b)=5/13,求cosb
1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?2.已知tana*tanb=13/7,tan[(a+b)/2]=(根号6)/2,求cos(a-b)3.已知a,b∈(0,π),且tan(a/2)=1/2,sin(a+b)=5/13,求cosb
1、tan2a=[tan(a+b)+tan(a-b)]/[1-tan(a+b)·tan(a-b)]=(3+2)/(1-3*2)=-1
∴sin4a=2tan2a/[1+(tan2a)^2]=2(-1)/(1+1)=-1
2、cos(a+b)={1-tan[(a+b)/2]^2}/{1+tan[(a+b)/2]^2}=(1-6/4)/(1+6/4)=-1/5
又 cos(a+b)=cosa·cosb-sina·sinb=-1/5
13/7=tana·tanb=sina·sinb/(cosa·cosb)
∴-1/5=cosa·cosb-sina·sinb=cosa·cosb-13/7cosa·cosb=-6/7cosa·cosb
∴cosa·cosb=7/30
∴cos(a-b)=cosa·cosb+sina·sinb=cosa·cosb+13/7cosa·cosb=20/7cosa·cosb=2/3
3、sina=2tan(a/2)/{1+[tan(a/2)]^2}=2*1/2/(1+1/4)=4/5
cosa={1-[tan(a/2)]^2}/{1+[tan(a/2)]^2}=(1-1/4)/(1+1/4)=3/5
又 sin(a+b)=5/13
∴cos(a+b)=±12/13
∴cosb=cos(a+b-a)=cos(a+b)·cosa+sin(a+b)·sina=56/65 或 -16/65