Kako pronaći točnu vrijednost COS (SIN ^ -1 4/5 + TAN ^ -1 5/12)?

Odgovor:

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

Obrazloženje:

pustiti #sin ^ (- 1) (4/5) = x # zatim

# 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 = tan ^ (- 1) (4/3) = sin ^ (- 1) = (4/5) #

Sada,

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

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

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

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

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

pustiti #tan ^ (- 1) (63/16) = A # zatim

# RarrtanA = 63/16 #

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

# RarrA = cos ^ (- 1) (16/65) = tan ^ (- 1) (63/16), #

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

Kako ćete pronaći točnu vrijednost grijeha (cos ^ -1 (sqrt5 / 5))?

Sin (cos ^ -1 (sqrt (5) / 5)) = (2sqrt (5)) / 5 Neka cos ^ -1 (sqrt (5) / 5) = A onda cosA = sqrt (5) / 5 i sinA = sqrt (1-cos ^ 2A) = sqrt (1- (sqrt (5) / 5) ^ 2) = (2sqrt (5)) / 5 rarrA = sin ^ -1 ((2sqrt (5)) / 5) Sada, grijeh (cos ^ -1 (sqrt (5) / 5)) = sin (sin ^ -1 ((2sqrt (5)) / 5)) = (2sqrt (5)) / 5

Kako ćete pronaći točnu vrijednost od 112,5 stupnjeva koristeći polumotnu formulu?

Tan (112.5) = - (1 + sqrt (2)) 112.5 = 112 1/2 = 225/2 NB: Ovaj kut leži u 2. kvadrantu. => Tan (112,5) = tan (225/5) = sin (225/2) / cos (225/2) = - sqrt ([sin (225/2) / cos (225/2)] ^ 2) = -sqrt (sin ^ 2 (225/2) / cos ^ 2 (225/2)) Kažemo da je negativan jer je vrijednost tan uvijek negativna u drugom kvadrantu! Zatim koristimo donju formulu za polu-ugao: sin ^ 2 (x / 2) = 1/2 (1-cosx) cos ^ 2 (x / 2) = 1/2 (1 + cosx) => tan (112,5) = -sqrt (sin ^ 2 (225/2) / cos ^ 2 (225/2)) = -sqrt ((1/2 (1-cos (225))) / (1/2 (1 + cos (225) ))))) = -sqrt ((1-cos (225)) / (1 + cos (225))) Primijetite da: 225 = 180 + 45 => cos (2

Kako pronaći točnu vrijednost tan [arc cos (-1/3)]?

Koristite trigonometrijski identitet tan (theta) = sqrt ((1 / cos ^ 2 (theta) -1)) Rezultat: tan [arccos (-1/3)] = boja (plava) (2sqrt (2)) Započni ostavljajući arccos (-1/3) da bude kut theta => arccos (-1/3) = theta => cos (theta) = - 1/3 To znači da sada tražimo tan (theta). identitet: cos ^ 2 (theta) + sin ^ 2 (theta) = 1 Podijelite sve obje strane cos ^ 2 (theta) da imate, 1 + tan ^ 2 (theta) = 1 / cos ^ 2 (theta) = > tan ^ 2 (theta) = 1 / cos ^ 2 (theta) -1 => tan (theta) = sqrt ((1 / cos ^ 2 (theta) -1)) Podsjetimo, ranije smo rekli da cos (theta) = -1 / 3 => tan (theta) = sqrt (1 / (- 1/3) ^ 2-1) = s