Tunjukkan bahawa (a ^ 2sin (B-C)) / (sinB + sinC) + (b ^ 2sin (C-A)) / (sinC + sinA) + (c ^ 2sin (A-B)) / (sinA + sinB)

Bahagian pertama

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

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

# = (4R ^ 2sin (pi- (B + C)) dosa (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) #

Begitu juga

Bahagian 2

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

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

Bahagian ke-3

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

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

Menambah tiga bahagian yang kita ada

Ungkapan yang diberikan #=0#

Jika 2sin theta + 3cos theta = 2 membuktikan bahawa 3sin theta - 2 cos theta = ± 3?

Sila lihat di bawah. Memandangkan rarr2sinx + 3cosx = 2 rarr2sinx = 2-3cosx rarr (2sinx) ^ 2 = (2-3cosx) ^ 2 rarr4sin ^ 2x = 4-6cosx + 9cos ^ 2x rarrcancel (4) -4cos ^ 2x = cancel (4) - 6cosx + 9cos ^ 2x rarr13cos ^ 2x-6cosx = 0 rarrcosx (13cosx-6) = 0 rarrcosx = 0,6 / 13 rarrx = 90 ° Sekarang, 3sinx-2cosx = 3sin90 ° -2cos90 ° =

Bagaimana anda membuktikan (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 * A + B) / 2) * cos ((AB) / 2)] ^ 2 = 4cos ^ 2 ((AB) / 2) [sin ^ 2 ((A + B) / 2) + B) / 2)] = 4cos ^ 2 ((AB) / 2) * 1 = 4cos ^ 2 ((AB) / 2) = RHS

(CosA + 2CosC) / (CosA + 2CosB) = SinB / SinC, Buktikan bahawa segitiga sama ada isosel atau betul bersudut?

Memandangkan rarr (cosA + 2cosC) / (cosA + 2cosB) = sinB / sinC rarrcosAsinB + 2sinB * cosB = cosAsinC + 2sinCcosC rarrcosAsinB + sin2B = cosAsinC + sin2C rarrcosA (sinB-sinC) + sin2B-sin2C = 0 rarrcosA [ 2) + 2 * sin (2B-2C) / 2) * cos ((2B + 2C) / 2)] = 0 rarrcosA [2sin ((BC (B + C) / 2)] + 2 * sin (BC) * cos (B + C)] = 0 rarrcosA [2sin ((BC) / 2) (Cos) (cos) (cos) (cos) (cos) (cos) (2) / 2) + 2cos ((BC) / 2)] = 0 Sama ada, cosA = 0 rarrA = 90 ^ @ atau, dosa ((BC) / 2) = 0 rarrB = C Oleh itu, segitiga sama ada isosceles atau kanan bersudut . Kredit pergi ke dk_ch tuan.