Poišči vrednost theta, če, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Odgovor:

# theta = pi / 3 # ali #60^@#

Pojasnilo:

V redu. Imamo:

# costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 #

Prezrimo # RHS # za zdaj.

# costheta / (1-sintheta) + costheta / (1 + sintheta) #

# (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sintheta) (1 + sintheta)) #

# (costheta ((1-sintheta) + (1 + sintheta))) / (1-sin ^ 2theta) #

# (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) #

# (2costheta) / (1-sin ^ 2theta) #

V skladu s pitagorejsko identiteto, # sin ^ 2theta + cos ^ 2theta = 1 #. Torej:

# cos ^ 2theta = 1-sin ^ 2theta #

Zdaj, ko to vemo, lahko pišemo:

# (2costheta) / cos ^ 2theta #

# 2 / costheta = 4 #

# costheta / 2 = 1/4 #

# costheta = 1/2 #

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

# theta = pi / 3 #, kdaj # 0 <= theta <= pi #.

V stopinjah, # theta = 60 ^ @ # kdaj # 0 ^ @ <= theta <= 180 ^ @ #

Odgovor:

# rarrcosx = 1/2 #

Pojasnilo:

Glede, # rarrcosx / (1-sinx) + cosx / (1 + sinx) = 4 #

#rarrcosx 1 / (1-sinx) + 1 / (1 + sinx) = 4 #

#rarrcosx (1 + preklic (sinx) + 1zaključi (-sinx)) / ((1-sinx) * (1 + sinx) = 4 #

#rarr (2cosx) / (1-sin ^ 2x) = 4 #

# rarrcosx / cos ^ 2x = 2 #

# rarrcosx = 1/2 #