2sinBsinC+cos2C=1+cos2A-cos2B
三角函数的问题 1+cos2A-cos2B-cos2C=2sinBsinC,求角A -
2cos²A-2cos²B+2sin²C=2sinBsinC cos²A-cos²B+sin² (A+B)=sinBsinC 2sinAcosAsinBcosB+2cos²Asin²B=sinBsinC 2cosA(sinAcosB+cosAsinB)-sinC=0 2cosAsinC-sinC=0 SinC(2cosA-1)=0 cosA=1/2 A=60 还有呢?
1+cos2A-cos2B-cos2C=2sinBsinC 1+1-2sin²A-1+2sin²B-1+2sin²C=2sinBsinC sin²B+sin²C-sin²A=sinBsinC sin²B+sin²C-sin²[180°-(B+C)]=sinBsinC sin²B+sin²C-sin²(B+C)=sinBsinC sin²B+等我继续说。
2cos²A-2cos²B+2sin²C=2sinBsinC cos²A-cos²B+sin² (A+B)=sinBsinC 2sinAcosAsinBcosB+2cos²Asin²B=sinBsinC 2cosA(sinAcosB+cosAsinB)-sinC=0 2cosAsinC-sinC=0 SinC(2cosA-1)=0 cosA=1/2 A=60 还有呢?
1+cos2A-cos2B-cos2C=2sinBsinC 1+1-2sin²A-1+2sin²B-1+2sin²C=2sinBsinC sin²B+sin²C-sin²A=sinBsinC sin²B+sin²C-sin²[180°-(B+C)]=sinBsinC sin²B+sin²C-sin²(B+C)=sinBsinC sin²B+等我继续说。
解:1+cos2A-cos2B-cos2C=2sinBsinC 2cos²A-1-2cos²B+1+2sin²C=2sinBsinC cos²A-cos²B+sin² (A+B)=sinBsinC cos²A-cos²B+sin²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC cos²A-cos²A等我继续说。
cos2A-cos2B=1-cos2C,,', 1-2sin^2 A 2sin^2 B-1=2sin^2 C ,',sin^2A sin^2C=sin^2B ','sinA=a/2R,sinB=b/2R,sinC=c/2R .'.a^2 c^2=b^2,三角形ABC为直角三角形,
在三角形ABC.角ABC的对边为abc.且1+cos2A-cos2B-cos2C=2sinBsinC...
解:1+cos2A-cos2B-cos2C=2sinBsinC 根据倍角公式cos2a=1-2sin²a可得:1+1-2sin²A-(1-2sin²B)-(1-2sin²C)=2sinBsinC化简,得:sin²A+sin²B+sin²C=sinBsinC 根据正玄定理sinB=bsinA/a ,sinc=csinA/a,可得:sin²A+b&#等我继续说。
简单分析一下,答案如图所示,
cos^2A+cos^2B+cos^2c=1 -
2cos^2Bcos^2C=2(sinBcosCcosBsinC)两边同除2cosBcosC得即为cosBosC=sinBsinC
解:(Ⅰ)由1+cos2A-cos2B-cos2C=2sinB·sinC,得 ,由正弦定理,得 ,由余弦定理,得 ,∵0<A<π,∴ 。(Ⅱ) ,由(Ⅰ)得, ,∴ ,∴ ,∵0<B< , ∴ ,令 ,即 时, 取得最大值 .
在△ABC中,若cos^2A+cos^2B+cos^2C=1,则△ABC的形状是? -
cos^2A+cos^2B+cos^2C=1 cos^2B+cos^2C=1-cos^2A 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^2C)=2(sinBcosCcosBsinC)cos^2Bcos^2C+cos^2Ccos^说完了。
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^2C)=2(sinBcosCcosBsinC)cos^2Bcos^2C+cos^2Ccos^2B=2(sinBcosCcosBsinC)即cosBcosC=sinBsinC 即tanBtanC=1等会说。