Vis at (a ^ 2sin (B-C)) / (sinB + sinC) + (b2sin (C-A)) / (sinC + sinA) + (c2sin (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) #

Tilsvarende

2. del

# = (B ^ 2sin (C-A)) / (SINC + sina) #

# = 4R ^ 2 (sinc-sina) #

3. del

# = (C ^ 2sin (A-B)) / (sina + sinB) #

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

Tilføjelse af tre dele, vi har

Det givne udtryk #=0#

Hvordan beviser du (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 + B) / 2)] = 4cos ^ 2 ((AB) / 2) * 1 = 4cos ^ 2 ((AB) / 2) = RHS

(CosA + 2CosC) / (CosA + 2CosB) = SinB / SinC, Bevis at trekanten er enten lige eller vinkelret?

Givet 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 ( ) / Cos (B + C) / 2)] + 2 * sin (BC) * cos (B + C)] = 0 rarrcosA [2sin ((BC) / 2) * cos ) / Cos (* BC) / 2) * cos ((BC) / 2)] = 0 rarr2cosA * sin ((BC) / 2) [cos ((B + C) / 2) + 2cos ((BC) / 2)] = 0 Enten, cosA = 0 rarrA = 90 ^ @ eller sin ((BC) / 2) = 0 rarrB = C Derfor er trekanten enten ensformet eller retvinklet . Kredit g

Bekræft at synd (A + B) + synd (A-B) = 2sinA sinB?

"se forklaring"> "ved hjælp af" farve (blå) "additionsformler til synd" • farve (hvid) (x) sin (A + -B) = sinAcosB + -cosAsinB rArrsin (A + B) = sinAcosB + cosAsinB rArrsin ) = sinAcosB-cosAsinB rArrsin (A + B) + sin (AB) = 2sinAcosB! = 2sinAsinBlarr "tjek dit spørgsmål"