Tunjukkan bahawa, (1 + cos theta + i * sin theta) ^ n + (1 + cos theta - i * sin theta) ^ n = 2 ^ (n + 1) * (cos theta / n * theta / 2)?

Jawapan:

Sila lihat di bawah.

Biarkan $1 + costheta + isintheta = r (cosalpha + isinalpha)$ , di sini $r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta)$

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

dan # tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / atau $alpha = theta / 2$

kemudian $1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha)$

dan kita boleh menulis $(1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n$ menggunakan teorem DE MOivre sebagai

$r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha)$

= $2r ^ ncosnalpha$

= $2 * 2 ^ ncos ^ n (theta / 2) cos ((ntheta) / 2)$

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