# rarrtan75 ° = μαύρισμα (45 + 30) #
# = (tan45 + tan30) / (1-tan45 * tan30) #
# = (1+ (1 / sqrt (3))) / (1- (1 / sqrt (3)) #
# = (sqrt (3) + 1) / (sqrt (3) -1) = 2 + sqrt (3)
# rarrtan52.5 = κούνια (90-37.5) = cot37.5 #
# rarrcot37.5 = 1 / (μαύρισμα (75/2)) #
# rarrtanx = (2tan (χ / 2)) / (1-tan ^ 2 (χ / 2)) #
# rarrtanx-tanx * tan ^ 2 (χ / 2) = 2tan (χ / 2) #
# rarrtanx * tan ^ 2 (x / 2) + 2tan (χ / 2) -tantx = 0 #
Είναι τετραγωνικό #tan (x / 2) # Ετσι, #rarrtan (x / 2) = (- 2 + sqrt (2 ^ 2-4 * tanx * (- tanx)) /
#rarrtan (x / 2) = (- 2 + sqrt (4 (1 + tan ^ 2x))) / (2 * tanx)
#rarrtan (x / 2) = (- 1 + sqrt (1 + tan ^ 2x)) / tanx #
Βάζοντας # x = 75 # παίρνουμε
#rarrtan (75/2) = (- 1 + sqrt (1 + tan ^ 2 (75)) / (tan75)
(2 + sqrt (3)) ^ 2)) / (2 + sqrt (3)) #
#rarrtan (75/2) = (- 1 + sqrt (1 + 4 + 4sqrt (3) +3)) / (2 + sqrt (3)
#rarrtan (75/2) = (- 1 + sqrt (8 + 4sqrt (3)) / (2 + sqrt (3)
(2 + sqrt (3)) + (1) (2 + 2) / (2 * sqrt (2 + sqrt (3) + 1) #
(2 * sqrt (2 + sqrt (3)) + 1)) / (2 * sqrt (3 + sqrt (2 + sqrt (3))) ^ 2-1 ^ 2) #
Αφήνω #sqrt (2 + sqrt (3)) = α #
# rarrcot37.5 = (2 * (2 * α + 1) + sqrt (3) * (2 * α + 1) / / (4 *
# rarrcot37.5 = (4a + 2 + 2sqrt (3) a + sqrt (3)) / (4 * (2 + sqrt (3)
# rarrcot37.5 = (4a + 2sqrt (3) α + α ^ 2) / (7 + 4sqrt (3)) * (7-4sqrt (3)
# rarrcot37.5 = 7 * (4a + 2sqrt (3) α + α ^ 2) -4sqrt (3) * (4a + 2sqrt (3)
# rarrcot37.5 = 28a + 14sqrt (3) a + 7a ^ 2-16sqrt (3) a-24a-4sqrt
# rarrcot37.5 = 7a ^ 2-4sqrt (3) a ^ 2 + 4a-2sqrt (3) α #
# rarrcot37.5 = α ^ 2 (7-4sqrt (3)) + 2 * a (2-sqrt (3)
# rrtrcot37.5 = (2 + sqrt (3)) (7-4sqrt (3)) + 2 * sqrt (2 + sqrt (3)) sqrt -sqrt (3))) #
# rarrcot37.5 = 2 * (7-4sqrt (3)) + sqrt (3) (7-4sqrt (3)) + sqrt (2 ^ 2 *
# rarrcot37.5 = 14-8sqrt (3) + 7sqrt (3) -12 + sqrt ((sqrt (6) -sqrt (2)
# rarrtan52.5 = 2-sqrt (3) + sqrt (6) -sqrt (2) #
Αποδείχθηκε.
Απάντηση:
Μικρότερη προσέγγιση …
Κανόνες που χρησιμοποιούνται: -
#color (κόκκινο) (ul (bar (| χρώμα (πράσινο) (sin2theta = 2 cdot sintheta cdot costheta)) |
# cos2theta = 2cos ^ 2theta-1 #
# => χρώμα (κόκκινο) (ul (bar (| χρώμα (μπλε) (2cos ^ 2theta = 1 + cos2theta)
Εξήγηση:
#tan (52.5 ^ @) #
# = αμαρτία (52.5 ^ @) / cos (52.5 ^ @) #
# = αμαρτία (105/2) ^ @ / cos (105/2) ^ @ #
# (2 cdot cos (105/2) ^ @ cdot cos (105/2) ^ @) / (2 cdot cos (105/2) ^ @ cdot cos (105/2) ^ @
# = αμαρτία (105/2 xx2) ^ @ / (2 cdot cos ^ 2 (105/2) ^ @ #
# = sin (105) ^ @ / (cos (105) ^ @ + 1) #
# = αμαρτία (60 ^ @ + 45 ^ @) / (cos (60 ^ + 45 ^ @) + 1)
# = (sin60 ^ @ cdot cos45 ^ @ cos60 ^ @ cdot sin45 ^ @) / (cos60 ^ @ cdot cos45 ^ @ -sin60 ^ @ cdot sin45 ^
# = (sqrt3 / 2 cdot 1 / sqrt2 + 1/2 cdot 1 / sqrt2) / (1/2 cdot 1 / sqrt2-sqrt3 / 2 cdot 1 / sqrt2 + 1 #
# = ((sqrt3 + 1) / (2sqrt2)) / ((1-sqrt3 + 2sqrt2) / (2sqrt2) #
# = (sqrt3 + 1) / (1-sqrt3 + 2sqrt2 #
= ((sqrt3 + 1) cdot (1 + 2sqrt2 + sqrt3)) / ((1 + 2sqrt2) ^ 2- (sqrt3)
# = (sqrt3 + 2sqrt6 + 3 + 1 + 2sqrt2 + sqrt3) / (1 + 4sqrt2 + 8-3) #
# = (2 (sqrt6 + sqrt3 + sqrt2 + 2)) / (6 + 4sqrt2) #
# = ((sqrt6 + sqrt3 + sqrt2 + 2)) / (3 + 2sqrt2) #
= ((3-2sqrt2) (sqrt6 + sqrt3 + sqrt2 + 2)) / (3 + 2sqrt2) (3-2sqrt2) #
# = (sqrt6-sqrt3-sqrt2 + 2) / 1 #
# = sqrt6-sqrt3-sqrt2 + 2 #
Ελπίζω να βοηθά …
Ευχαριστώ…
# tan105 ^ = tan (60 ^ @ + 45 ^ @) #
(= tan60 ^ @ + tan45 ^ @) / (1-tan60 ^ tan45 ^ @) #
# => tan105 ^ @ = (sqrt3 + 1) / (1-sqrt3 * 1) #
# => tan105 ^ @ = (1 + sqrt3) / (1-sqrt3) #
# => tan105 ^ @ = - ((sqrt3 + 1) (sqrt3-1)) / (sqrt3-1) ^ 2 #
# => tan105 ^ @ = - (3-1) / (4-2sqrt3) #
= = (2tn52.5 ^) / (1-tan ^ 2 52.5 ^) = - 1 / (2-sqrt3) #
Αφήνω #tan52.5^@=x#
Τώρα
# (2χ) / (1-χ ^ 2) = - 1 / (2-sqrt3) #
# => x ^ 2-2 (2-sqrt3) x-1 = 0 #
(2 (2-sqrt3) + sqrt (4 (2-sqrt3) ^ 2 + 4)) / 2 #
Οπως και # 52.5^@in "Πρώτο τεταρτημόριο -η ρίζα παραμελήθηκε" #
(2-sqrt3) + 2sqrt ((2-sqrt3) ^ 2 + 1)) / 2 #
# => x = (2-sqrt3) + sqrt ((2-sqrt3) ^ 2 + 1) #
# => x = (2-sqrt3) + sqrt (8-4sqrt3) #
# => x = (2-sqrt3) + sqrt (2 (4-2sqrt3) #
# => x = (2-sqrt3) + sqrt (2 (sqrt3-1) ^ 2) #
# => x = 2-sqrt3 + sqrt2 (sqrt3-1) #
# => x = 2-sqrt3 + sqrt6-sqrt2 #
# => x = sqrt6-sqrt3-sqrt2 + 2 #
