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

Odgovor:

# = Sin (theta) / (1 + cos (theta)) *

Obrazloženje:

# (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)) *

Nađi vrijednost theta, ako, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 ili 60 ^ @ U redu. Imamo: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Zanemarimo RHS za sada. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sinteta) (1 + sinteta)) (costheta ((1-sintheta) ) + (1 + sinteta))) / (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Pitagorejski identitet, sin ^ 2teta + cos ^ 2tea = 1. Dakle: cos ^ 2theta = 1-sin ^ 2theta Sada kada znamo da, možemo pisati: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ - 1 (1/

Dokaz: - sin (7 theta) + sin (5 theta) / sin (7 theta) -sin (5 theta) =?

(sin7x + sin5x) / (sin7x-sin5x) = tan6x * cotx rarr (sin7x + sin5x) / (sin7x-sin5x) = (2sin ((7x + 5x) / 2) * cos ((7x-5x) / 2) ) / (2sin ((7x-5x) / 2) * cos ((7x + 5x) / 2) = (sin6x * cosx) / (sinx * cos6x) = (tan6x) / tanx = tan6x * cottx

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)?

Pogledajte dolje. Neka 1 + costheta + isintheta = r (cosalpha + isinalpha), ovdje r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2) ) -2) = 2cos (theta / 2) i tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) ili alfa = theta / 2 zatim 1 + costheta-isintheta = r (cos (-alfa) + isin (-alfa)) = r (cosalpha-isinalpha) i možemo pisati (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n koristeći DE MOivreov teorem kao r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 * 2 ^ n