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

Svar:

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

Forklaring:

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

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

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

= 2 (2 + 2 sin (theta) + 2 cos (theta) + 2 sin (theta) cos (theta))

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

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

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

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

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

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

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

Find værdien af theta, hvis Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 eller 60 ^ @ Okay. Vi har: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Lad os ignorere RHS for nu. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) ((1-sintheta) (1 + sintheta) ) + (1 + sintheta)) / (1-sin2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin2theta) (2costheta) / (1-sin ^ 2theta) den pythagoranske identitet, sin ^ 2theta + cos ^ 2theta = 1. Så: cos ^ 2theta = 1-sin ^ 2theta Nu da vi ved det, kan vi skrive: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ 1 (1/2) theta = pi / 3, n

Vis at, (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)?

Se nedenfor. Lad 1 + costheta + isintheta = r (cosalpha + isinalpha), her r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2 ) -2) = 2cos (theta / 2) og tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) eller alfa = theta / 2 derefter 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) og vi kan skrive (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n ved anvendelse af DE MOivre's sætning som rnn (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosna

Forenkle udtrykket :? (Sin ^ 2 (pi / 2 + a) -cos ^ 2 (a-pi / 2)) / (tg ^ 2 (pi / 2 + a) -ctg ^ 2 (a-pi / 2))

(sin ^ 2 (pi / 2 + alfa) -koser2 (alfa-pi / 2)) / (tan ^ 2 (pi / 2 + alfa) -cot ^ 2 (alfa-pi / 2)) = ^ 2 (pi / 2 + alfa) -koser2 (pi / 2-alfa)) / (tan ^ 2 (pi / 2 + alfa) -cot ^ 2 (pi / 2-alfa)) = (cos ^ 2 (alfa) -sin ^ 2 (alfa)) / (cot ^ 2 (alfa) -tan ^ 2 (alfa)) = (cos ^ 2 (alfa) -sin ^ 2 (alfa)) / (cos ^ 2 ) / sin ^ 2 (alfa) -sin ^ 2 (alfa) / cos ^ 2 (alfa)) = (cos ^ 2 (alfa) -sin ^ 2 (alfa)) / ((cos ^ 4 (alfa) ^ 4 (alfa)) / (sin ^ 2 (alfa) cos ^ 2 (alfa)) = (cos ^ 2 (alfa) -sin ^ 2 (alfa)) / (cos ^ 4 (alfa) -sin ^ 4 (alfa)) xx (sin ^ 2 (alfa) cos ^ 2 (alfa)) / 1 = (cos ^ 2 (alfa) -sin ^ 2 (alfa)) / ((cos ^ 2 (alfa) -si