Pasimplehin (1- cos theta + sin theta) / (1+ cos theta + sin theta)?

Sagot:

# = sin (theta) / (1 + cos (theta)) #

Paliwanag:

# (1-cos (theta) + sin (theta)) / (1 + cos (theta) + sin (theta)) #

(1 + cos (theta) + sin (theta)) * (1 + cos (theta) + sin (theta)) /

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

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

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

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

(1 + cos (theta)) / (1 + cos (theta)) - (1/2) (cos ^ 2 (theta)) / ((1 + cos (theta)) (1 + kasalanan (theta))) #

(1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1-sin ^ 2 (theta)) / ((1 + cos (theta)) (1 + kasalanan (theta))) #

(1 + sin (theta)) / (1 + cos (theta)) - (1/2) ((1-sin (theta)) * ((1 + cos (theta)) (1 + kasalanan (theta))) #

(1 + cos (theta)) / (1 + cos (theta)) - (1/2) (1-sin (theta)) /

# = sin (theta) / (1 + cos (theta)) #

Hanapin ang halaga ng theta, kung, Cos (theta) / 1 - kasalanan (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 or 60 ^ @ Okay. Mayroon kaming: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Huwag pansinin ang RHS para sa ngayon. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) (1-sintheta) ) (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Ayon sa ang Pythagorean Identity, sin ^ 2theta + cos ^ 2theta = 1. Kaya: cos ^ 2theta = 1-sin ^ 2theta Ngayon na alam natin na, maaari nating isulat: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ 1 (1/2) theta = pi / 3,

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)?

Mangyaring tingnan sa ibaba. (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) (theta / 2) o alpha = theta / 2 pagkatapos 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) at maaari naming isulat (1 + costheta + ^ n + (1 + costheta-isintheta) ^ gamit ang teorem 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)

Pasimplehin ang expression :? (sin ^ 2 (pi / 2 + alpha) -cos ^ 2 (alpha-pi / 2)) / (tg ^ 2 (pi / 2 + alpha) -ctg ^ 2 (alpha-pi / 2))

(sin ^ 2 (pi / 2 + alpha) -cos ^ 2 (alpha-pi / 2)) / (tan ^ 2 (pi / 2 + alpha) -cot ^ 2 (alpha-pi / 2) 2 (pi / 2 + alpha) -cos ^ 2 (pi / 2-alpha)) / (tan ^ 2 (pi / 2 + alpha) -cot ^ 2 (pi / 2-alpha) (alpha) -in ^ 2 (alpha)) / (cot ^ 2 (alpha) -tan ^ 2 (alpha)) = (cos ^ 2 (alpha) -sin ^ 2 (alpha)) / (cos ^ 2 (alpha ) / sin ^ 2 (alpha) -sin ^ 2 (alpha) / cos ^ 2 (alpha)) = (cos ^ 2 (alpha) -sin ^ 2 (alpha)) / ((cos ^ 4 (alpha) ^ (Alpha)) / (sin ^ 2 (alpha) cos ^ 2 (alpha))) = (cos ^ 2 (alpha) -sin ^ 2 (alpha)) / (cos ^ 4 (alpha) (alpha)) xx (sin ^ 2 (alpha) cos ^ 2 (alpha)) / 1 = (cos ^ 2 (alpha) -sin ^ 2 (alpha)) / ((cos ^