1、已知A、B是锐角,且sinA=√5/5,sinB=√10-查字典问答网
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来自程银波的问题

  1、已知A、B是锐角,且sinA=√5/5,sinB=√10/10.求A+B的值2、若sinA+cosA=tanA(0<A<π/2),这A所在区间()A(0,π/6)B(π/6,π/4)C(π/4,π/3)D(π/3,π/2)3、已知sin²C/sin²A+tan(A-B)/tanA

  1、已知A、B是锐角,且sinA=√5/5,sinB=√10/10.求A+B的值

  2、若sinA+cosA=tanA(0<A<π/2),这A所在区间()

  A(0,π/6)B(π/6,π/4)C(π/4,π/3)D(π/3,π/2)

  3、已知sin²C/sin²A+tan(A-B)/tanA=1求证tanAtanB=tan²C

  4、化简sin²(A+π/6)+sin²(A-π/6)-sin²A

  5、求值①cos20ºcos40ºcos60ºcos80º②sin66ºsin42ºsin6ºsin78º

  ③tan67.5º-tan22.5º④cosπ/12·cos5π/12

  6、化简(1+sinA-cosA)/(1+sinA+cosA)+(1+sinA+cosA)/(1+sinA-cosB)

  7、已知sin(π/4+A)sin(π/4-A)=1/6,且(π/2<A<π)求sin4A

  8、已知tanA=1/7,tanB=1/3且A、B都为锐角,求A+2B

  9、求值sin²20º+sin²80º+√3sin20ºcos80º

  10、求值[2sin50º+sin10º(1+√3tan10º)]·√(2sin²80º)

  11、若sin(π/6-A)=1/3则cos(2π/3+2A)=?

  12、已知tan(π/4+A)=3求sin2A-2cos²A的值

  13、在△ABC中,已知B=π/3则tanA/2+tanC/2+√3tanA/2tanC/2=?

  14、化简√(1+cos10º)

  15、已知cos(A-π/6)+sinA=4√3/5,求sin(A+7π/6)的值

  16、3sinB=sin(2A-B),求证tan(A+B)=2tanA

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2020-03-04 20:03
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  1.∵A、B是锐角

  ∴cosA=√[1-(sinA)^2]=2√5/5,cosB=√[1-(sinB)^2]=3√10/10

  cos(A+B)=cosAcosB-sinAsinB=√2/2

  A+B=45º

  2.sinA+cosA=√2sin(A+π/4)=tanA

  ∵0<A<π/2

  ∴π/4<A+π/4<3π/4

  则√2/2<sin(A+π/4)<1

  ∴1<tanA<√2

  π/4<tanA<arctan√2,约为(π/4,π/3),选C

  3.tan(A-B)=(tanA-tanB)/(1+tanA*tanB)

  1-[(tanA-tanB)/(1+tanA*tanB)]/tanA=(sinC)^2/(sinA)^2

  [(tanA)^2*tanB+tanB]/[tanA(1+tanA*tanB)]=(sinC)^2/(sinA)^2

  tanB*(secA)^2/tanA(1+tanA*tanB)=(sinC)^2/(sinA)^2,(其中1+tan^2A=sec^2A)

  tanB*(secA)^2*(sinA)^2=tanA(1+tanA*tanB)*(sinC)^2

  tanB*tanA=(1+tanA*tanB)*(sinC)^2,(两边除以tanA)

  (tanB*tanA+1)-1=(1+tanA*tanB)*(sinC)^2

  1-1/(1+tanA*tanB)=(sinC)^2

  1-(sinC)^2=1/(1+tanA*tanB)

  (cosC)^2=1/(1+tanA*tanB)

  1/(secC)^2=1/(1+tanA*tanB)

  (secC)^2=1+tanA*tanB

  (secC)^2-1=tanA*tanB

  (tanC)^2=tanA*tanB

  4.原式=[sinAcos(π/6)+cosAsin(π/6)]^2+[sinAcos(π/6)-cosAsin(π/6)]^2-(sinA)^2

  =[(√3/2)sinA+(1/2)cosA]^2+[(√3/2)sinA-(1/2)cosA]^2-(sinA)^2

  =3(sinA)^2/2+(cosA)^2/2-(sinA)^2=1/2*[(sinA)^2+(cosA)^2]

  =1/2

  5.①原式=2sin20°cos20°cos40°cos80°/4sin20°

  =sin40°cos40°cos80°/4sin20°

  =sin80°cos80°/8sin20°

  =sin160°/16sin20°

  =sin(180°-20°)/16sin20°

  =1/16

  ②原式=sin(90°-24°)sin(90°-48°)sin6°sin(90°-12°)

  =cos24°cos48°sin6°cos12°

  =2sin6°cos6°cos12°cos24°cos48°/2cos6°

  =sin12°cos12°cos24°cos48°/2cos6°

  =sin24°cos24°cos48°/4cos6°

  =sin48°cos48°/8cos6°

  =sin96°/16cos6°

  =1/16

  ③原式=sin67.5°/cos67.5°-sin22.5°/cos22.5°

  =cos22.5°/sin22.5°-sin22.5°/cos22.5°

  =[(cos22.5°)^2-(sin22.5°)^2]/(sin22.5°*cos22.5°)

  =cos45°/[(1/2)*sin45°]

  =2

  ④原式=1/2*[cos(5π/12+π/12)+cos(5π/12-π/12)]

  =1/2*[cos(π/2)+cos(π/3)]

  =1/2*[0+1/2]

  =1/2*1/2

  =1/4

  6.原式=[(1+sinA-cosA)^2+(1+sinA+cosA)^2]/(1+sinA+cosA)(1+sinA-cosA)

  展开,整理后=4(1+sinA)/2sinA(sinA+1)

  =2/sinA

  7.sin(π/4+A)sin(π/4-A)=(-1/2){cos[(π/4+A)+(π/4-A)]-cos[(π/4+A)-(π/4-A)]}

  =(-1/2)[cos(π/2)-cos2A]=(1/2)cos2A=1/6

  cos2A=1/3

  ∵π/2<A<π

  ∴π<2A<2π

  sin2A=-√[1-(cos2A)^2]=-2√2/3

  sin4A=2sin2A*cos2A=-4√2/9

  8.tan2B=2tanB/1-(tanB)^2=3/4

  tan(A+2B)=(tanA+tan2B)/(1-tanAtan2B)=1

  ∵A、B都为锐角

  ∴A+2B∈(0,270°)

  A+2B=45°或A+2B=225°

  9.原式=(1-cos40°)/2+(1+cos160°)/2+√3/2*(sin100°-sin60°)

  =1+(cos160°-cos40°)/2+√3/2*sin100°-3/4

  =1-sin[(160°+40°)/2]*sin[(160°-40°)/2]+√3/2*sin100°-3/4

  =1-sin100°sin60°+√3/2*sin100°-3/4

  =1/4

  10.原式=[2sin50°+sin10°(cos10°+√3sin10°)/cos10°]×√2sin80°

  =[2sin50°cos10°+2sin10°(cos60°cos10°+sin60°sin10°)×√2sin80°/cos10°

  =2[sin50°cos10°+sin10°cos(60°-10°)]×√2

  =2sin(50°+10°)×√2

2020-03-04 20:07:26

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