Odgovor:
Glej spodaj.
Pojasnilo:
Let # 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 # alpha = theta / 2 #
potem # 1 + costheta-isintheta = r (cos (-alfa) + isin (-alpha)) = r (cosalpha-isinalpha) #
in lahko pišemo # (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n # z uporabo DE MOivreovega izreka kot
# r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) #
= # 2r ^ ncosnalpha #
= # 2 * 2 ^ ncos ^ n (theta / 2) cos ((ntheta) / 2) #
= # 2 ^ (n + 1) cos ^ n (theta / 2) cos ((ntheta) / 2) #
