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

odpoveď:

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

vysvetlenie:

# (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-cos2 (theta) / (1 + cos2 (theta) + sin2 (theta) +2 sin (theta) +2 cos (theta) + 2 sin (theta) cos (theta)) #

# = ((1 + sin (theta)) 2-cos2 (theta) / (2 + 2 sin (theta) +2 cos (theta) + 2 sin (theta) cos (theta)) #

# = ((1 + sin (theta)) 2-cos2 (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)) #

Zistite hodnotu theta, ak, Cos (theta) / 1-sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 alebo 60 ^ Okay. Máme: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Teraz budeme ignorovať RHS. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sintheta) (1 + sintheta) (costheta ((1-sintheta) ) + (1 + sintheta)) / (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Pythagorean Identity, sin ^ 2theta + cos ^ 2theta = 1. Takže: cos ^ 2theta = 1-sin ^ 2theta Teraz, keď to vieme, môžeme napísať: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 t

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

Ak frac {sin theta} {x} = frac {cos theta] {y} potom heta theta - cos theta = pm frac {x - y} {sqrt {x ^ 2 + y ^ 2}} frac { sin theta} {x} = frac {cos theta] {y} frac {heta theta} {cos theta} = frac {x} {y} heta = x / y Je to ako pravý trojuholník s opakom x a susediace y tak cos theta = frac {pm y} {sqrt {x ^ 2 + y ^ 2} sin theta = theta = theta = theta = theta theta = teta theta cos theta (theta - 1) = frac {pm y} {sqrt {x ^ 2 + y ^ 2}} (x / y -1) heta theta - cos theta = pm frac {x - y } {sqrt {x ^ 2 + y ^ 2}}

Ukážte, že (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)?

Pozri nižšie. Nech 1 + costheta + isintheta = r (cosalpha + isinalpha), tu r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2 ) -2) = 2cos (theta / 2) a tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) alebo alfa = theta / 2 potom 1 + costheta-isintheta = r (cos (-alfa) + isín (-alfa)) = r (cosalpha-isinalpha) a môžeme písať (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n pomocou vety DE MOivre ako r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 * 2 ^