Če je sin theta + cos theta = p, kaj je sin ^ 2 theta + cos ^ 4theta v smislu p?

Odgovor:

# 1 - ((p ^ 2-1) / 2) ^ 2 #

Pojasnilo:

# (sintheta + costheta) ^ 2 = 1 + 2sinthetacostheta = p ^ 2 #

tako

#sinthetacostheta = (p ^ 2-1) / 2 #

zdaj

# sin ^ 2theta + cos ^ 4theta = sin ^ 2theta + (1-sin ^ 2theta) cos ^ 2theta = 1-sin ^ 2thetacos ^ 2theta #

in vse skupaj

# sin ^ 2theta + cos ^ 4theta = 1 - ((p ^ 2-1) / 2) ^ 2 #

Kako izražate cos (4theta) v smislu cos (2theta)?

Cos (4theta) = 2 (cos (2theta)) ^ 2-1 Začnite tako, da nadomestite 4-ti z 2theta + 2theta cos (4theta) = cos (2theta + 2theta) Poznavanje cos (a + b) = cos (a) cos ( b) -sin (a) sin (b), nato cos (2theta + 2theta) = (cos (2theta)) ^ 2- (sin (2theta)) ^ 2 vedoč, da (cos (x)) ^ 2+ (sin ( x)) ^ 2 = 1, potem (sin (x)) ^ 2 = 1- (cos (x)) ^ 2 rarr cos (4theta) = (cos (2theta)) ^ 2- (1- (cos (2theta)) ) ^ 2) = 2 (cos (2theta)) ^ 2-1

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 izražate cos theta - cos ^ 2 theta + sec theta v smislu greha theta?

Sqrt (1-sin ^ 2 theta) - (1-sin ^ 2 theta) + 1 / sqrt (1-sin ^ 2 theta) samo še bolj poenostavi, če je potrebno. Iz danih podatkov: Kako izražate cos theta-cos ^ 2 theta + sec theta v smislu sin-theta? Rešitev: iz osnovnih trigonometričnih identitet Sin ^ 2 theta + Cos ^ 2 theta = 1 sledi cos theta = sqrt (1-sin ^ 2 theta) cos ^ 2 theta = 1-sin ^ 2 theta tudi sec theta = 1 / cos theta torej cos theta-cos ^ 2 theta + sec theta sqrt (1-sin ^ 2 theta) - (1-sin ^ 2 theta) + 1 / sqrt (1-sin ^ 2 theta) Bog blagoslovi ... upam razlaga je koristna.