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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=180° ∴A+B = 180°- C ∴sin(A+B)=sin(180°-C)= sinC

sin(A+B)=sinC成立 cos(A+B)=-cosC

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

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

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

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...

根据正弦定理,(a+b)/a= sinB/(sinB -sinA) =(sinA+sinB)/sinA ∴sinA·sinB = (sinB+sinA)(sinB-sinA) = 2sin[(B+A)/2]·cos[(B-A)/2]·2·cos[(B+A)/2]·sin[(B-A)] =sin(B-A)·sin(B+A) =sinC·sin(B-A) cos(A-B)+cosC=1-cos2C 即 2sinA·sinB = 2(sin...

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