Nađi vrijednost theta, ako, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Odgovor:

# Theta = pi / 3 # ili #60^@#

Obrazloženje:

U redu. Imamo:

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

Zanemarimo # RHS # zasad.

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

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

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

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

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

Prema pitagorejskom identitetu, # Grijeh ^ 2 theta + cos ^ 2 theta = 1 #, Tako:

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

Sada kada to znamo, možemo pisati:

# (2costheta) / cos ^ 2 theta #

# 2 / costheta = 4 #

# Costheta / 2 = 1/4 #

# Costheta = 1/2 #

# Theta = cos ^ 1 (1/2) #

# Theta = pi / 3 #, kada # 0 <theta <pi #.

U stupnjevima, # Theta = 60 ^ '# kada # 0 ^ '<theta <180 ^' #

Odgovor:

# Rarrcosx = 1/2 #

Obrazloženje:

S obzirom, # Rarrcosx / (1-sinx) + cosx / (1 + sinx) = 4 #

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

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

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

# Rarrcosx / cos ^ 2x = 2 #

# Rarrcosx = 1/2 #