Poenostavite (1 - cos theta + sin theta) / (1+ cos theta + sin theta)?

Odgovor:

# = sin (theta) / (1 + cos (theta)) #

Pojasnilo:

# (1-cos (theta) + sin (theta)) / (1 + cos (theta) + sin (theta)) #

# = (1-cos (theta) + sin (theta)) * (1 + cos (theta) + sin (theta)) / (1 + cos (theta) + sin (theta)) ^ 2 #

# = ((1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / (1 + cos ^ 2 (theta) + sin ^ 2 (theta) +2 sin (theta) +2 cos (theta) + 2 sin (theta) cos (theta)) #

# = ((1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / (2 + 2 sin (theta) +2 cos (theta) + 2 sin (theta) cos (theta)) #

# = ((1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / (2 (1 + cos (theta)) + 2 sin (theta) (1 + cos (theta)) #

# = (1/2) ((1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / ((1 + cos (theta)) (1 + sin (theta)) #

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (cos ^ 2 (theta)) / ((1 + cos (theta)) (1 + sin (theta))) #

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1-sin ^ 2 (theta)) / ((1 + cos (theta)) (1 + sin (theta))) #

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) ((1-sin (theta)) * (1 + sin (theta))) / ((1 + cos (theta)) (1 + sin (theta))) #

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1-sin (theta)) / (1 + cos (theta)) #

# = sin (theta) / (1 + cos (theta)) #

Poišči vrednost theta, če, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 ali 60 ^ @ OK. Dobili smo: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Zdaj ignoriramo RHS. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sintheta) (1 + sintheta)) (costheta ((1-sintheta) ) + (1 + sinteta)) / (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Pitagorejska identiteta, sin ^ 2teta + cos ^ 2tea = 1. Torej: cos ^ 2theta = 1-sin ^ 2theta Zdaj, ko vemo, da lahko, pišemo: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ - 1 (1/2)

Pokažite, da (1 + cos theta + i * sin theta) ^ n + (1 + cos theta - i * sin theta) ^ n = 2 ^ (n + 1) * (cos theta / 2) ^ n * cos ( n * theta / 2)?

Glej spodaj. Naj bo 1 + costheta + isintheta = r (cosalpha + isinalpha), tukaj r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2) ) -2) = 2cos (theta / 2) in tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) ali alfa = theta / 2, nato 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) in lahko napišemo (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n z uporabo DE MOivrejevega izreka kot r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 *

Poenostavite (-i sqrt 3) ^ 2. kako to poenostavite?

-3 Izvirno funkcijo lahko zapišemo v njeni razširjeni obliki, kot je prikazano (-isqrt (3)) (- isqrt (3)), obravnavamo kot spremenljivko in ker je negativni čas negativ enak pozitivnemu, in kvadratni koren krat kvadratni koren iste številke je preprosto, da je število, dobimo spodaj enačbo i ^ 2 * 3 Ne pozabite, da i = sqrt (-1) in deluje s pravilo kvadratnega korena prikazano zgoraj, lahko poenostavimo, kot je prikazano spodaj -1 * 3 Zdaj je stvar aritmetične -3 In tam je vaš odgovor :)