水淼cos冰雪2b
@云石2502：sin(A - B)=0,求证cos(A - 2B)=cosA - 作业帮
沙家15695919315…… [答案] cos(A-2B)=cos(2A-2B-A)=cos(2A-2B)cosA+sin(2A-2B)sinA=cos2(A-B)cosA+sin2(A-B)sinA =cos2(A-B)cosA=[1-2sin(A-B)^2]cosA=cosA.
@云石2502：在三角形abc中,若cos^2A+cos^2B>1+cos^2C,则三角形ABC的形状 - 作业帮
沙家15695919315…… [答案] cos^2A+cos^2B>1+cos^2C 1-sin^2A+1-sin^2B>1+1-sin^2C sin^2C>sin^2A+sin^2B a/sinA=b/sinB=c/sinC c^2>a^2+b^2 cosC=(a^2+b^2-c^2)/2ab<0,所以c为钝角 三角形ABC是钝角三角形.
@云石2502：(sinα - sinβ)/sin(α - β)=a,(cosα - cosβ)/sin(α+β)=b,求sin(α - β)的值 -
沙家15695919315…… (sinα-sinβ)/sin(α-β)=a (sinα-sinβ)=a*sin(α-β) 2cos[(α+β)/2]*sin[(α-β)/2]=2a*sin[(α-β)/2]*cos...
@云石2502：一道高一数学题 -
沙家15695919315…… tan(π/4+b)=(tanπ/4+tanb)/(1-tanπ/4tanb)=(1+tanb)/(1-tanb)=3所以:tanb=1/2sin2b-2cos^2 b=sin2b-cos2b-1=2tanb/(1+tan^2b)-(1-tan^2b)/(1+tan^2b)-1=1/(1+1/4)-(1-1/4)/(1+1/4)-1=4/5-3/5-1=-4/5
@云石2502：三角形ABC中,a+c=2b,A - C=π/3,求sinB. -
沙家15695919315…… 由正弦定理: a:b:c=sinA:sinB:sinC => sinA + sinC =2sin((A-C)/2)*cos((A+C)/2) =cos((π-B)/2) =sin(B/2) =sinB =2sin(B/2)cos(B/2) => cos(B/2)=1/2 B/2=π/3 B=2π/3 => sinB=根号3/2
@云石2502：(1+2sinbcosb)/cos^2b - sin^2b= - 3的解题过程 - 作业帮
沙家15695919315…… [答案] (1+2sinbcosb)/cos^2b-sin^2b =(sinb+cosb)^2/[(cosb+sinb)(cosb-sinb)] =(sinb+cosb)/(cosb-sinb)=-3 所以sinb=2cosb,即tanb=2
@云石2502：在△ABC中,若等试1+cos^2C=cos^2A+cos^2B成立,求证△ABC为直角三角形 - 作业帮
沙家15695919315…… [答案] 原题应该是;1— cos^2C=cos^2A+cos^2B成立,求证△ABC为直角三角形 cos^2B+cos^2C=sin^2A cos^2B+cos^2C=sin^2(B+C) cos^2B+cos^2C=sin^2Bcos^2C+2(sinBcosCcosBsinC)+cos^2Bsin^2C cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2...
@云石2502：cos2A - cos 2B= 的解法 -
沙家15695919315…… cos2A-cos 2B=cos[(A+B)+(A-B)]-cos[(A+B)-(A-B)]=cos(A+B)cos(A-B)-sin(A+B)sin(A-B)-[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]=cos(A+B)cos(A-B)-sin(A+B)sin(A-B)-cos(A+B)cos(A-B)-sin(A+B)sin(A-B)=-2sin(A+B)sin(A-B)
@云石2502：cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明 - 作业帮
沙家15695919315…… [答案] 左边=sin(A+B)sin(B-A)+sin²C =sin(180-C)sin(B-A)+sin²C =sinCsin(B-A)+sin²C =sinC[sin(B-A)+sinC] =sinC[sin(B-A)+sin(B+A)] =sinC(sinBcosA-cosBsinA+sinBcosA+cosBsinA) =sinC*2sinBcosA=右边 命题得证
沙家15695919315…… [答案] cos(A-2B)=cos(2A-2B-A)=cos(2A-2B)cosA+sin(2A-2B)sinA=cos2(A-B)cosA+sin2(A-B)sinA =cos2(A-B)cosA=[1-2sin(A-B)^2]cosA=cosA.
@云石2502：在三角形abc中,若cos^2A+cos^2B>1+cos^2C,则三角形ABC的形状 - 作业帮
沙家15695919315…… [答案] cos^2A+cos^2B>1+cos^2C 1-sin^2A+1-sin^2B>1+1-sin^2C sin^2C>sin^2A+sin^2B a/sinA=b/sinB=c/sinC c^2>a^2+b^2 cosC=(a^2+b^2-c^2)/2ab<0,所以c为钝角 三角形ABC是钝角三角形.
@云石2502：(sinα - sinβ)/sin(α - β)=a,(cosα - cosβ)/sin(α+β)=b,求sin(α - β)的值 -
沙家15695919315…… (sinα-sinβ)/sin(α-β)=a (sinα-sinβ)=a*sin(α-β) 2cos[(α+β)/2]*sin[(α-β)/2]=2a*sin[(α-β)/2]*cos...
@云石2502：一道高一数学题 -
沙家15695919315…… tan(π/4+b)=(tanπ/4+tanb)/(1-tanπ/4tanb)=(1+tanb)/(1-tanb)=3所以:tanb=1/2sin2b-2cos^2 b=sin2b-cos2b-1=2tanb/(1+tan^2b)-(1-tan^2b)/(1+tan^2b)-1=1/(1+1/4)-(1-1/4)/(1+1/4)-1=4/5-3/5-1=-4/5
@云石2502：三角形ABC中,a+c=2b,A - C=π/3,求sinB. -
沙家15695919315…… 由正弦定理: a:b:c=sinA:sinB:sinC => sinA + sinC =2sin((A-C)/2)*cos((A+C)/2) =cos((π-B)/2) =sin(B/2) =sinB =2sin(B/2)cos(B/2) => cos(B/2)=1/2 B/2=π/3 B=2π/3 => sinB=根号3/2
@云石2502：(1+2sinbcosb)/cos^2b - sin^2b= - 3的解题过程 - 作业帮
沙家15695919315…… [答案] (1+2sinbcosb)/cos^2b-sin^2b =(sinb+cosb)^2/[(cosb+sinb)(cosb-sinb)] =(sinb+cosb)/(cosb-sinb)=-3 所以sinb=2cosb,即tanb=2
@云石2502：在△ABC中,若等试1+cos^2C=cos^2A+cos^2B成立,求证△ABC为直角三角形 - 作业帮
沙家15695919315…… [答案] 原题应该是;1— cos^2C=cos^2A+cos^2B成立,求证△ABC为直角三角形 cos^2B+cos^2C=sin^2A cos^2B+cos^2C=sin^2(B+C) cos^2B+cos^2C=sin^2Bcos^2C+2(sinBcosCcosBsinC)+cos^2Bsin^2C cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2...
@云石2502：cos2A - cos 2B= 的解法 -
沙家15695919315…… cos2A-cos 2B=cos[(A+B)+(A-B)]-cos[(A+B)-(A-B)]=cos(A+B)cos(A-B)-sin(A+B)sin(A-B)-[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]=cos(A+B)cos(A-B)-sin(A+B)sin(A-B)-cos(A+B)cos(A-B)-sin(A+B)sin(A-B)=-2sin(A+B)sin(A-B)
@云石2502：cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明 - 作业帮
沙家15695919315…… [答案] 左边=sin(A+B)sin(B-A)+sin²C =sin(180-C)sin(B-A)+sin²C =sinCsin(B-A)+sin²C =sinC[sin(B-A)+sinC] =sinC[sin(B-A)+sin(B+A)] =sinC(sinBcosA-cosBsinA+sinBcosA+cosBsinA) =sinC*2sinBcosA=右边 命题得证
