Nađite vrijednost grijeha (a + b) ako je tan a = 4/3 i cot b = 5/12, 0 ^ stupnjeva

Odgovor:

#sin (a + b) = 56/65 #

S obzirom, # tana = 4/3 i cotb = 5/12 #

# Rarrcota = 3/4 #

# Rarrsina = 1 / CSCA = 1 / sqrt (1 + krevetić ^ 2a) = 1 / sqrt (1+ (3/4) ^ 2) = 4/5 #

# Rarrcosa = sqrt (1-sin ^ 2a) = sqrt (1- (4/5) ^ 2) = 3/5 #

# Rarrcotb = 5/12 #

# Rarrsinb = 1 / cscb = 1 / sqrt (1 + krevetić ^ 2b) = 1 / sqrt (1+ (5/12) ^ 2) = 12/13 #

# Rarrcosb = sqrt (1-sin ^ 2b) = sqrt (1- (12/13) ^ 2) = 5/13 #

Sada, #sin (a + b) = sina * cosb + cosa * # sinb

#=(4/5)(5/13)+(3/5)*(12/13)=56/65#

#sin (a + b) = 56/65 #

Ovdje, # 0 ^ circ <boja (ljubičasta) (a) <90 ^ circ => I ^ (st) kvadrant => boja (plava) (sve, fn.> 0. #

# 0 ^ circ <boja (ljubičasta) (b) <90 ^ circ => I ^ (st) kvadrant => boja (plava) (sve, fn.> 0 #

Tako, # 0 ^ circ <boja (ljubičasta) (a + b) <180 ^ circ => I ^ (st) i II ^ (nd) kvadrant #

# => boja (plava) (sin (a + b)> 0 #

Sada, # Tana = 4/3 => seca = + sqrt (1 + tan ^ 2a) = sqrt (1 + 16/9) = 5/3 #

#:. boja (crvena) (cosa) = 1 / seca = boja (crvena) (3/5 #

# => Boja (crvena) (sina) + = sqrt (1-cos ^ 2a) = sqrt (1-9 / 25) = boja (crvena) (4/5 #

Također, # Cotb = 5/12 => cscb = + sqrt (1 + krevetić ^ 2b) = sqrt (1 + 25/144) = 13/12 #

#:. boja (crvena) (sinb) = 1 / cscb = boja (crvena) (12/13 #

# => Boja (crvena) (cosb) + = sqrt (1-sin ^ 2b) = sqrt (1-144 / 169) = boja (crvena) (5/13 #

Stoga, #sin (a + b) = + sinacosb cosasinb #

# => Sin (a + b) = 4 / 5xx5 / 13 + 3 / 5xx12 / 13 #

#sin (a + b) = 20/65 + 36/65 = 56/65 #