Paano mo patunayan ang 1 / (1 + kasalanan (theta)) + 1 / (1-kasalanan (theta)) = 2sec ^ 2 (theta)?

Sagot:

Tingnan sa ibaba

Paliwanag:

LHS = kaliwang bahagi, RHS = kanang bahagi

LHS# = 1 / (1 + sin theta) + 1 / (1-sin theta) #

# = (1-sin ang theta + 1 + sin theta) / ((1 + sin theta) (1-sin theta)) #-> Karaniwang denominador

# = (1-cancelsin theta + 1 + cancelsin theta) / ((1 + sin theta) (1-sin theta)) #

# = 2 / (1-sin ^ 2x) #

# = 2 / cos ^ 2x #

# = 2 * 1 / cos ^ 2x #

# = 2sec ^ 2x #

# = RHS #

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,

Paano mo i-convert ang r = 2sec (theta) sa form na cartesian?

X = 2 r = 2 / costheta rcostheta = 2 rcostheta = x = 2 x = 2

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)