Ipakita na, (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)?

Sagot:

Mangyaring tingnan sa ibaba.

Hayaan # 1 + costheta + isintheta = r (cosalpha + isinalpha) #, dito # r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) #

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

at # tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2) o # alpha = theta / 2 #

pagkatapos # 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) #

at maaari naming isulat # (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n # gamit ang teorama ng DE MOivre bilang

# 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) #