(CosA + 2CosC)/(CosA + 2CosB)= SinB / SinC,证明三角形是等腰角还是直角?

(CosA + 2CosC)/(CosA + 2CosB)= SinB / SinC,证明三角形是等腰角还是直角?
Anonim

特定 #rarr(COSA + 2cosC)/(COSA + 2cosB)= SINB / SINC#

#rarrcosAsinB + 2sinB * =的CoSb + cosAsinC#2sinCcosC

#rarrcosAsinB + sin2B = cosAsinC + sin2C#

#rarrcosA(SINB-SINC)+ sin2B-sin2C = 0#

#rarrcosA 2sin((BC)/ 2)* cos((B + C)/ 2) + 2 * SIN((2B-2C)/ 2)* cos((2B + 2C)/ 2) = 0 #

#rarrcosA 2sin((B-C)/ 2)* cos((B + C)/ 2) + 2 * SIN(B-C)* COS(B + C) = 0#

#rarrcosA 2sin((BC)/ 2)* cos((B + C)/ 2) + COSA * 2 * 2 * SIN((BC)/ 2)* cos((BC)/ 2) = 0 #

#rarr2cosA * SIN((B-C)/ 2)cos((B + C)/ 2)+ 2COS((B-C)/ 2) = 0#

要么, #COSA = 0# #rarrA = 90 ^ @#

要么, #sin((B-C)/ 2)= 0# #rarrB = C#

因此,三角形是等腰或直角。信用证转到dk_ch先生。