Kako dokazati greh (theta + phi) / cos (theta-phi) = (tantheta + tanphi) / (1 + tanthetatanphi)?

Odgovor:

Oglejte si spodnji dokaz

Pojasnilo:

Potrebujemo

#sin (a + b) = sinacosb + sinbcosa #

#cos (a-b) = cosacosb + sinasinb #

Zato, # LHS = sin (theta + phi) / cos (theta-phi) #

# = (sinthetacosphi + costhetasinphi) / (costhetacosphi + sinthetasinphi) #

Delitev z vsemi izrazi z# costhetacosphi #

# = ((sinthetacosphi) / (costhetacosphi) + (costhetasinphi) / (costhetacosphi)) / ((costhetacosphi) / (costhetacosphi) + (sinthetasinphi) / (costhetacosphi)) #

# = (sintheta / costheta + sinphi / cosphi) / (1 + sintheta / costheta * sinphi / cosphi) #

# = (tantheta + tanphi) / (1 + tanthetatanphi) #

# = RHS #

# QED #

Odgovor:

Glejte Razlago

Pojasnilo:

Let

# y = sin (theta + phi) / cos (theta-phi) #

# y = (sinthetacosphi + costhetasinphi) / (costhetacosphi + sinthetasinphi) #

Delitev s #cos theta #, # y = (tanthetacosphi + sinphi) / (cosphi + tanthetasinphi) #

Delitev s # cosphi #, # y = (tantheta + tanphi) / (1 + tanthetatanphi) #

zato se je izkazalo.

Odgovor:

# "glej razlago" #

Pojasnilo:

# "z uporabo" barve (modre) "trigonometrične identitete" #

# • barva (bela) (x) sin (x + y) = sinxcosy + cosxsiny #

# • barva (bela) (x) cos (x-y) = cosxcosy + sinxsiny #

# "upoštevajte levo stran" #

# = (sinthetacosphi + costhetasinphi) / (costhetacosphi + sinthetasinphi) #

# "razdelimo izraze na števec / imenovalec s" costhetacosphi #

# "in preklic skupnih dejavnikov" #

# = ((sinthetacosphi) / (costhetacosphi) + (costhetasinphi) / (costhetacosphi)) / ((costhetacosphi) / (costhetacosphi) + (sinthetasinphi) / (costhetacosphi)) = ((sintheta) / costheta + sinphi / cosphi) / (1 + sintheta / costhetaxxsinphi / cosphi #

# = (tantheta + tanphi) / (1 + tanthetatanphi) #

# = "desna stran" rArr "preverjeno" #

Pokažite, da cos²π / 10 + cos²4π / 10 + cos² 6π / 10 + cos²9π / 10 = 2. Malo sem zmeden, če naredim Cos²4π / 10 = cos² (π-6π / 10) & cos²9π / 10 = cos² (π-π / 10), bo postal negativen kot cos (180 ° - theta) = - costheta v drugi kvadrant. Kako naj dokazujem vprašanje?

Glej spodaj. LHS = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 ((6pi) / 10) + cos ^ 2 ((9pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi- (4pi) / 10) + cos ^ 2 (pi- (pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) = 2 * [cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [cos ^ 2 (pi / 2- (4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [sin ^ 2 ((4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * 1 = 2 = RHS

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 *

Kako rešiti greh 3 theta = 1?

Theta = pi / 6 + 2 / 3npi kjer je n celo število. Poznavanje, da greh (pi / 2) = 1 Poznavanje, da greh (x + 2pi) = sin (x), potem 3theta = pi / 2 + 2npi, kjer je n celo število rarr theta = (pi / 2 + 2npi) / 3 = pi / 6 + 2 / 3npi