Pokažite, da (a ^ 2sin (B-C)) / (sinB + sinC) + (b ^ 2sin (C-A)) / (sinC + sinA) + (c ^ 2sin (A-B)) / (sinA + sinB) = 0?

1. del

# (a ^ 2sin (B-C)) / (sinB + sinC) #

# = (4R ^ 2sinAsin (B-C)) / (sinB + sinC) #

# = (4R ^ 2sin (pi- (B + C)) sin (B-C)) / (sinB + sinC) #

# = (4R ^ 2sin (B + C) sin (B-C)) / (sinB + sinC) #

# = (4R ^ 2 (sin ^ 2B-sin ^ 2C)) / (sinB + sinC) #

# = 4R ^ 2 (sinB-sinC) #

podobno

2. del

# = (b ^ 2sin (C-A)) / (sinC + sinA) #

# = 4R ^ 2 (sinC-sinA) #

3. del

# = (c ^ 2sin (A-B)) / (sinA + sinB) #

# = 4R ^ 2 (sinA-sinB) #

Dodajmo tri dele, ki jih imamo

Navedeni izraz #=0#

Kako dokazujete (cosA + cosB) ^ 2 + (sinA + sinB) ^ 2 = 4 * cos ^ 2 ((A-B) / 2)? 2)?

LHS = (cosA + cosB) ^ 2 + (sinA + sinB) ^ 2 = [2 * cos ((A + B) / 2) * cos ((AB) / 2)] ^ 2+ [2 * sin ( A + B) / 2) * cos ((AB) / 2)] ^ 2 = 4cos ^ 2 ((AB) / 2) [sin ^ 2 ((A + B) / 2) + cos ^ 2 ((A + B) / 2)] = 4cos ^ 2 ((AB) / 2) * 1 = 4cos ^ 2 ((AB) / 2) = RHS

(CosA + 2CosC) / (CosA + 2CosB) = SinB / SinC, Dokaži, da je trikotnik enakokraka ali pravokoten?

Glede na rarr (cosA + 2cosC) / (cosA + 2cosB) = sinB / sinC rarrcosAsinB + 2sinB * 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 ((BC ) / 2) * cos ((B + C) / 2)] + 2 * sin (BC) * cos (B + C)] = 0 rarrcosA [2sin ((BC) / 2) * cos ((B + C) ) / 2)] + cosA * 2 * 2 * sin ((BC) / 2) * cos ((BC) / 2)] = 0 rarr2cosA * sin ((BC) / 2) [cos ((B + C) / 2) + 2cos ((BC) / 2)] = 0 Bodisi, cosA = 0 rarrA = 90 ^ @ ali, sin ((BC) / 2) = 0 rarrB = C Torej je trikotnik en

Preverite, ali greh (A + B) + sin (A-B) = 2sinA sinB?

"glej razlago"> "z uporabo" barvnih (modrih) "dodatkovnih formul za sin" • barva (bela) (x) sin (A + -B) = sinAcosB + -cosAsinB rArrsin (A + B) = sinAcosB + cosAsinB rArrsin (AB ) = sinAcosB-cosAsinB rArrsin (A + B) + sin (AB) = 2sinAcosB! = 2sinAsinBlarr "preverite svoje vprašanje"