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)?

Odgovor:

Pogledajte dolje.

pustiti # 1 + + costheta isintheta = r (cosalpha + isinalpha) #, ovdje # R = sqrt ((1 + costheta) ^ 2 + sin ^ 2 theta) = sqrt (2 + 2costheta) #

= #sqrt (2 + 4cos ^ 2 (theta / 2) -2) = 2cos (theta / 2) *

i # Tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) # ili # Alfa-theta / 2 #

zatim # 1 + costheta-isintheta = r (cos (alfa) + ISIN (alfa)) = f (cosalpha-isinalpha) #

i možemo pisati # (1 + + costheta isintheta) ^ n + (1 + costheta-isintheta) ^ n # koristeći DE MOivreov teorem kao

# R ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) #

= # 2r ^ ncosnalpha #

= # 2 x 2 ^ dočasnici ^ n (theta / 2) cos ((ntheta) / 2) *

= # 2 ^ (n + 1) cos ^ n (theta / 2) cos ((ntheta) / 2) *