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

Vastaus:

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

Selitys:

# (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 (teta) +2 cos (teta) + 2 sin (theta) cos (theta)) #

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

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

Etsi theta-arvo, jos, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 tai 60 ^ @ Okei. Meillä on: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Noudattakaa nyt RHS: ää. costeta / (1-sinteta) + costeta / (1 + sintetaatti) (costeta (1 + sintheta) + costeta (1-sintheta)) / ((1-sintheta) (1 + sintheta)) (costheta ((1-sintheta) ) + (1 + sintheta))) / (1-sin ^ 2theta) (Costeta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Pythagorean identiteetti, sin ^ 2theta + cos ^ 2theta = 1. Joten: cos ^ 2theta = 1-sin ^ 2theta Nyt kun tiedämme, että voimme kirjoittaa: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 =

Sin theta / x = cos theta / y sitten sin theta - cos theta =?

Jos frac {sin theta} {x} = frac {cos theta] {y}, sitten sin theta - cos theta = pm frac {x - y} {sqrt {x ^ 2 + y ^ 2}} frac { sin theta} {x} = frac {cos theta] {y} frac {sin theta} {cos theta} = frac {x} {y} heta = x / y Tämä on kuin oikea kolmio, jossa on vastakkainen x ja vierekkäin y niin cos theta = frac {y} {sqrt {x ^ 2 + y ^ 2} sin theta = theta-theta, teta - cos theta = tan theta cos-theta - cos theta = cos-theta (teta - 1) = frac {y} {sqrt {x ^ 2 + y ^ 2}} (x / y-1) sin theta - cos theta = pm frac {x - y } {sqrt {x ^ 2 + y ^ 2}}

Näytä, että (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)?

Katso alla. Olkoon 1 + costeta + isintheta = r (cosalpha + isinalpha), tässä r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2 ) -2) = 2cos (theta / 2) ja tanalpha = sintheta / (1 + costeta) == (2sin (teta / 2) cos (teta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) tai alfa = theta / 2, sitten 1 + costeta-isintheta = r (cos (alfa) + isiini (-alfa)) = r (cosalpha-isinalpha) ja voimme kirjoittaa (1 + costeta + isintheta) ^ n + (1 + costeta-isintheta) ^ n käyttäen DE MOivren teoriaa r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha