Kako ocenjujete sek ((5pi) / 12)?

Odgovor:

# 2 / (sqrt (2 - sqrt3)) #

Pojasnilo:

sek = 1 / cos. Ocenite cos ((5pi) / 12)

Krog Trig enot in lastnost komplementarnih lokov da ->

#cos ((5pi) / 12) = cos ((6pi) / 12 - (pi) / 12) = cos (pi / 2 - pi / 12) = sin (pi / 12) #

Poiščite sin (pi / 12) z uporabo identitete trigona:

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

#cos (pi / 6) = sqrt3 / 2 = 1 - 2sin ^ 2 (pi / 12) #

# 2sin ^ 2 (pi / 12) = 1 - sqrt3 / 2 = (2 - sqrt3) / 2 #

# sin ^ 2 (pi / 12) = (2 - sqrt3) / 4 #

#sin (pi / 12) = (sqrt (2 - sqrt3)) / 2 # --> #sin (pi / 12) # je pozitiven.

Končno, #sec ((5pi) / 12) = 2 / (sqrt (2 - sqrt3)) #

Odgovor lahko preverite z uporabo kalkulatorja.

Odgovor:

#sec ((5pi) / 12) = sqrt6 + sqrt2 #

Pojasnilo:

#sec x = 1 / cosx #

#sec ((5pi) / 12) = 1 / cos ((5pi) / 12) #

# (5pi) / 12 = pi / 4 + pi / 6 #-> Razdeli v sestavljeni argument

# = 1 / cos (pi / 4 + pi / 6) #

-> uporaba #cos (A + B) = cosAcosB-sinAsinB #

# = 1 / (cos (pi / 4) cos (pi / 6) -sin (pi / 4) sin (pi / 6)) #

# = 1 / ((sqrt2 / 2) (sqrt3 / 2) - (sqrt2 / 2) (- 1/2)) #

# = 1 / (sqrt6 / 4 -sqrt2 / 4) = 1 / ((sqrt6-sqrt2) / 4) = 4 / (sqrt6-sqrt2) #

# = 4 / (sqrt6-sqrt2) * (sqrt6 + sqrt2) / (sqrt6 + sqrt2) #

# = (4 (sqrt6 + sqrt2)) / (6-2) = (4 (sqrt6 + sqrt2)) / 4 #

# = sqrt6 + sqrt2 #