Kesin COS değerini nasıl bulabilirim (SIN ^ -1 4/5 + TAN ^ -1 5/12)?

Cevap:

#rarrcos (sin ^ (- 1) (4/5) + tan ^ (- 1) (5/12)) = 16/65 #

Açıklama:

let #sin ^ (- 1) (4/5) = x # sonra

# Rarrsinx = 4/5 #

# Rarrtanx = 1 / cotx = 1 / (sqrt (CSC ^ 2x-1)) = 1 / (sqrt ((1 / SiNx) ^ 2-1)) = 1 / (sqrt ((1 / (4/5)) ^ 2-1)) = 4/3 #

# Rarrx = kahve renkli ^ (- 1) (4/3) = sin ^ (- 1) = (4/5) #

Şimdi,

#rarrcos (sin ^ (- 1) (4/5) + tan ^ (- 1) (5/12)) #

# = Cos (tan ^ (- 1) (4/3) + tan ^ (- 1) (5/12)) #

# = Cos (tan ^ (- 1) ((4/3 + 5/12) / (1- (4/3) * (5/12)))) #

# = Cos (tan ^ (- 1) ((63/36) / (16/36))) #

# = Cos (tan ^ (- 1) (63/16)) #

let #tan ^ (- 1) (63/16) bir # = sonra

# RarrtanA = 63/16 #

# RarrcosA = 1 / seCA = 1 / sqrt (1 + tan ^ 2A) = 1 / sqrt (1 + (63/16) ^ 2) = 16/65 #

# RarrA = cos ^ (- 1) (- 1) (63/16) # (16/65) açık kahve renkli ^ =

#rarrcos (sin ^ (- 1) (4/5) + tan ^ (- 1) (5/12)) cos = (tan ^ (- 1) (63/16)) cos (= cos ^ (- 1) (16/65)) 16/65 # =