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为什么sin(A+B)=sinC,和Cos(A+B)=%COSC

在一个三角形中,ABC的和为180度

∵A+B+C=π,∴B+C=π-A,∴cos(B+C)=cos(π-A)=-cos A.故选C.

sin(A+B)=sin(180_C)=sinC, 先求sinA和sinC,再求cosB=cos(180_A_B)=_cos(A+B)=_(cosAcosB_sinAsinB)

∵A、B、C是三角形的三个内角∴A+B=π-C对于A,cos(A+B)=cos(π-C)=-cosC,故A错对于B,sin(A+B)=sin(π-C)=sinC,故B对对于C,tan(A+B)=tan(π-C)=-tanC,故C错对于D,sin A+B 2 = cos C 2 ,故D错故选B

sin(B+C)=sinBcosC+sinCcosB sin(B-C)=sinBcosC-sinCcosB cos(B+C)=cosBcosC-sinCsinB sin(B-C)=cosBcosC+sinCsinB sin(π/2-A)=cosA cos(π/2-A)=sinA

(1)∵sinC+cosC=1?sinC2∴2sinC2cosC2+1?2sin2C2=1?sinC2∴2sinC2cosC2?2sin2C2=?sinC2∴2sin2C2?2sinC2cosC2=sinC2∴2sinC2(sin C2?cosC2)=sinC2∴sinC2?cos C2=12∴sin2C2?sinC+cos2C2=14∴sinC=34(2)由sinC2?cosC2=12>0得π4<c2<π2即π...

(Ⅰ)∵sin22C+sin2C?sinC+cos2C=1,∴4sin2Ccos2C+2sin2CcosC+1-2sin2C=1,整理得:2cos2C+cosC-1=0,即cosC=12,则C=60°;(Ⅱ)由余弦定理可知:cosC=a2+b2-c22ab=(a+b)2-2ab-c22ab=12,∴25-2ab-72ab=12,即ab=6,∴S△ABC=12absinC=332.

2.cos(A+B)+cosC =cos(180-c)+cosc =-cosc+cosc =0

解答:∵由△内角和=180°,∴A+B+C=180°,∴A+B=180°-C,∴sin﹙A+B﹚=sin﹙180-C﹚=sinC。∴cos﹙A+B﹚=cos﹙180-C﹚=-cosC

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