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

odpoveď:

# Theta = pi / 3 # alebo #60^@#

vysvetlenie:

Poriadku. Máme:

# Costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 #

Ignorujme to # RHS # na Teraz.

# 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) #

Podľa 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 #

# Theta = cos ^ -1 (1/2) #

# Theta = pi / 3 #, kedy # 0 <= theta <= pi #.

V stupňoch, # Theta = 60 ^ @ # kedy # 0 ^ '<= theta <= 180 ^ @ #

odpoveď:

# Rarrcosx = 1/2 #

vysvetlenie:

Vzhľadom k tomu, # Rarrcosx / (1-sinx) + cosx / (1 + sinx) = 4 #

#rarrcosx 1 / (1-sinx) + 1 / (1 + sinx) = 4 #

#rarrcosx (1 + zrušiť (sinx) + 1cancel (-sinx)) / ((1-sinx) * (1 + sinx) = 4 #

#rarr (2cosx) / (1-sin ^ 2x) = 4 #

# Rarrcosx / cos ^ 2x = 2 #

# Rarrcosx = 1/2 #

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

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

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 ^