Cevap:
Lütfen aşağıya bakın.
Açıklama:
Alıyoruz
# LHS = ten rengi 20 ^ circ + tan80 ^ circ + tan140 ^ circ #
#color (beyaz) (ASÇ) = tan20 ^ Circ + tan (60 ^ Circ + 20 ^ devir daim) + tan (120 ^ Circ + 20 ^ devir daim) #
#color (beyaz) (LHS) #=# Tan20 ^ Circ + (tan60 ^ Circ + tan20 ^ devir daim) / (1-tan60 ^ circtan20 ^ devir daim) + (tan120 ^ Circ + tan20 ^ devir daim) / (1-tan120 ^ circtan20 ^ devir daim) #
Kimys. #color (blue) (tan60 ^ circ = sqrt3, tan120 ^ circ = -sqrt3 ve tan20 ^ circ = t #
# LHS = t + (SQRT3 + T) / (1-sqrt3t) + (- SQRT3 + T) / (1 + sqrt3t) #
#color (beyaz) (ASÇ) = t + {(SQRT3 + T) (1 + sqrt3t) + (- SQRT3 + T) (1-sqrt3t)) / ((1-sqrt3t) (1 + sqrt3t)) #
#color (beyaz) (ASÇ) = t + (SQRT3 + 3t + t + sqrt3t ^ 2-SQRT3 + 3t + T-sqrt3t ^ 2) / (1-3t ^ 2) #
#color (beyaz) (ASÇ) = t + (8t) / (1-3t ^ 2) #
#color (beyaz) (ASÇ) = (t-3-t ^ 3 + 8t) / (1-3t ^ 2) #
#color (beyaz) (ASÇ) = (9t-3-t ^ 3) / (1-3t ^ 2) #
#color (beyaz) (ASÇ) = 3: (3t-t ^ 3) / (1-3t ^ 2) towhere, renkli (mavi) (t = tan20 ^ devir daim #
#color (beyaz) (LHS) = 3 (3tan20 ^ circ-tan ^ 3 20 ^ circ) / (1-3tan ^ 2 20 ^ circ) #
#color (beyaz) (ASÇ) = 3: tan3 (20 ^ Circ) toApply (2) # için # Teta = 20 ^ circ #
# LHS = 3tan60 ^ circ #
# LHS = 3sqrt3 = RHS #
Not:
# (1) tan (A + B) = (tanA + tanB) / (1-tanAtanB) #
# (2) tan3theta = (3tantheta-kahve renkli ^ 3theta) / (1-3tan ^ 2teta) #
# LHS = tan20 + tan80 + tan140 #
# = Tan20 + tan80 + tan (180-40) #
# = tan20 + tan80-tan 40 #
# = tan20 + sin 80 / cos 80-sin 40 / cos 40 #
# = sin 20 / cos 20+ (sin 80cos 40-cos 80sin 40) / (cos 80cos 40) #
# = (günah 20cos 80cos 40 + günah 40cos 20) / (cos 20cos 80cos 40) #
Şimdi bu ifadenin paydası
# = çünkü 20cos 80cos 40 #
# = (4 * 2sin 20cos 20cos 40cos 80) / (8sin 20) #
# = (2 * 2sin 40cos 40cos 80) / (8sin 20) #
# = (2sin 80cos80) / (8sin 20) #
# = (günah 160) / (8sin 20) #
# = (günah (180-20)) / (8sin 20) #
# = (günah 20) / (8sin 20) #
#=1/8#
bundan dolayı
# LHS = 8 (günah 20cos 80cos 40 + günah 40cos 20) #
# = 4sin 20 * (2cos 80cos 40) + 4 * 2sin 40cos 20 #
# = 4sin 20 (cos 120 + cos 40) +4 (sin 60 + sin 20) #
# = 4sin 20 (-1 / 2 + cos 40) +4 (sqrt3 / 2 + sin 20) #
# = - 2sin 20 + 4sin 20cos 40 + 2sqrt3 + 4sin 20 #
# = 4sin 20cos 40 + 2sqrt3 + 2sin 20 #
# = 2 (günah 60-günah 20) + 2sqrt3 + 2sin 20 #
# = 2 (sqrt3 / 2-sin 20) + 2sqrt3 + 2sin 20 #
# = sqrt3-2sin 20 + 2sqrt3 + 2sin 20 #
# = 3sqrt3 #
Anwer kullanan komik bir yaklaşım # 3sqrt3 # Verilen.
LHS'yi bildiğimiz şekilde yazabiliriz # sqrt3 = tan 60 #
# LHS = tan 20 + tan 80 + tan 140 #
# = 3sqrt3 + (tan 20-tan 60) + (tan 80-tan 60) + (tan 140-tan 60) #
# = 3sqrt3 + (tan 20-tan 60) + (tan 80-tan 60) + (tan (180-40) -tan 60) #
# = 3sqrt3 + (tan 20-tan 60) + (tan 80-tan 60) - (tan 40 + tan 60) #
# = 3sqrt3 + (sin 20 / cos 20-sin 60 / cos60) + (sin 80 / cos 80-sin 60 / cos60) - (sin 40 / cos40 + sin 60 / cos60) #
# = 3sqrt3-sin (60 - 20) / (cos 20cos60) + sin (80-60) / (cos 80cos60) -sin (60 + 40) / (cos40cos60) #
# = 3sqrt3- (2sin 40) / cos 20+ (2sin 20) / cos 80- (2sin 100) / cos 40 #
# = 3sqrt3- (4sin 20cos 20) / cos 20+ (4sin 10 cos 10) / gün 10- (4sin 40cos 40) / cos 40 #
# = 3sqrt3-4sin 20 + 4cos 10-4sin 40 #
# = 3sqrt3-4 (günah 20 + günah 40) + 4cos 10 #
# = 3sqrt3-4 (2 sin 30cos1 0) + 4cos 10 #
# = 3sqrt3-4 (2 * 1/2 * cos1 0) + 4cos 10 #
# = 3sqrt3-4cos 10 + 4cos 10 #
# = 3sqrt3 #
Cevap:
Aşağıdaki açıklama
Açıklama:
#, X = tan20 + tan80 + tan140 #
=# Sin20 / cos20 + sin80 / cos80 + tan (180-40) #
=# (Cos80 * sin20 + sin80 * cos20) / (cos80 * cos20) -tan40 #
=#sin (80 + 20) / (cos80 * cos20) -sin40 / cos40 #
=# Sin100 / (cos80 * cos20) -sin40 / cos40 #
=# Sin80 / (cos80 * cos20) -sin40 / cos40 #
=# (Sin80 * cos40-cos80 * sin40 * cos20) / (cos80 * cos40 * cos20) #
=# (Sin20 * (8sin80 * cos40-8cos80 * sin40 * cos20)) / (8cos80 * cos40 * cos20 * sin20) #
=# (Sin20 * (4sin120 + 4sin40-4cos20 * (sin120-sin40))) / (4cos80 * cos40 * sin40) #
=# (Sin20 * (4sin120 + 4sin40-4sin120 * cos20 + 4sin40 * cos20)) / (2cos80 * sin80) #
=# (Sin20 * (4sin60 + 4sin40-4sin60 * cos20 + 4sin40 * cos20)) / (sin160) #
=# (Sin20 * (4sin60 + 4sin40-2sin80-2sin40 + 2sin60 + 2sin20)) / (sin20) #
=# 6sin60 + 2sin40-2sin80 + 2sin20 #
=# 3sqrt3 + 2sin20- (2sin80-2sin40) #
=# 3sqrt3 + 2sin20-4cos60 * sin20 #
=# 3sqrt3 + 2sin20-2sin20 #
=# 3sqrt3 #
