Chứng tỏ rằng tan (52,5 °) = sqrt6 - sqrt3 - sqrt2 + 2?

Chứng tỏ rằng tan (52,5 °) = sqrt6 - sqrt3 - sqrt2 + 2?
Anonim

# rarrtan75 ° = tan (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 = cũi (90-37,5) = cot37,5 #

# rarrcot37.5 = 1 / (tan (75/2)) #

# rarrtanx = (2tan (x / 2)) / (1-tan ^ 2 (x / 2)) #

# rarrtanx-tanx * tan ^ 2 (x / 2) = 2tan (x / 2) #

# rarrtanx * tan ^ 2 (x / 2) + 2tan (x / 2) -tanx = 0 #

Nó là bậc hai trong #tan (x / 2) # Vì thế, #rarrtan (x / 2) = (- 2 + sqrt (2 ^ 2-4 * tanx * (- tanx))) / (2 * tanx) #

#rarrtan (x / 2) = (- 2 + sqrt (4 (1 + tan ^ 2x))) / (2 * tanx) #

#rarrtan (x / 2) = (- 1 + sqrt (1 + tan ^ 2x)) / tanx #

Đưa # x = 75 # chúng tôi nhận được

#rarrtan (75/2) = (- 1 + sqrt (1 + tan ^ 2 (75))) / (tan75) #

#rarrtan (75/2) = (- 1 + sqrt (1+ (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)) #

# rarr1 / tan (75/2) = (2 + sqrt (3)) / (2 * sqrt (2 + sqrt (3)) - 1) * (2 * sqrt (2 + sqrt (3)) + 1) / (2 * sqrt (2 + sqrt (3) +1) #

# rarrcot37.5 = (2 * (2 * sqrt (2 + sqrt (3)) + 1) + sqrt (3) * (2 * sqrt (2 + sqrt (3)) + 1)) / ((2 * sqrt (2 + sqrt (3))) ^ 2-1 ^ 2) #

Để cho #sqrt (2 + sqrt (3)) = a #

# rarrcot37.5 = (2 * (2 * a + 1) + sqrt (3) * (2 * a + 1)) / ((4 * (2 + sqrt (3))) ^ 2-1) #

# rarrcot37.5 = (4a + 2 + 2sqrt (3) a + sqrt (3)) / ((4 * (2 + sqrt (3)) - 1) #

# rarrcot37.5 = (4a + 2sqrt (3) a + a ^ 2) / (7 + 4sqrt (3)) * (7-4sqrt (3)) / (7-4sqrt (3)) #

# rarrcot37.5 = 7 * (4a + 2sqrt (3) a + a ^ 2) -4sqrt (3) * (4a + 2sqrt (3) a + a ^ 2) #

# rarrcot37.5 = 28a + 14sqrt (3) a + 7a ^ 2-16sqrt (3) a-24a-4sqrt (3) a ^ 2 #

# rarrcot37.5 = 7a ^ 2-4sqrt (3) a ^ 2 + 4a-2sqrt (3) a #

# rarrcot37.5 = a ^ 2 (7-4sqrt (3)) + 2 * a (2-sqrt (3)) #

# rarrcot37.5 = (2 + sqrt (3)) (7-4sqrt (3)) + 2 * sqrt (2 + sqrt (3)) * sqrt ((2-sqrt (3))) * sqrt ((2) -sqrt (3))) #

# rarrcot37.5 = 2 * (7-4sqrt (3)) + sqrt (3) (7-4sqrt (3)) + sqrt (2 ^ 2 * (2-sqrt (3))) #

# rarrcot37.5 = 14-8sqrt (3) + 7sqrt (3) -12 + sqrt ((sqrt (6) -sqrt (2)) ^ 2) #

# rarrtan52.5 = 2-sqrt (3) + sqrt (6) -sqrt (2) #

Chứng minh.

Câu trả lời:

Cách tiếp cận nhỏ hơn …

Quy tắc được sử dụng: -

#color (đỏ) (ul (thanh (| màu (xanh) (sin2theta = 2 cdot sintheta cdot costheta)) | #

# cos2theta = 2cos ^ 2theta-1 #

# => màu (đỏ) (ul (thanh (| màu (xanh) (2cos ^ 2theta = 1 + cos2theta)) | #

Giải trình:

#tan (52,5 ^ @) #

# = sin (52,5 ^ @) / cos (52,5 ^ @) #

# = sin (105/2) ^ @ / cos (105/2) ^ @ #

# = (2 cdot sin (105/2) ^ @ cdot cos (105/2) ^ @) / (2 cdot cos (105/2) ^ @ cdot cos (105/2) ^ @ #

# = sin (105/2 xx2) ^ @ / (2 cdot cos ^ 2 (105/2) ^ @ #

# = sin (105) ^ @ / (cos (105) ^ @ + 1) #

# = sin (60 ^ @ + 45 ^ @) / (cos (60 ^ @ + 45 ^ @) + 1) #

# = (sin60 ^ @ cdot cos45 ^ @ + cos60 ^ @ cdot sin45 ^ @) / (cos60 ^ @ cdot cos45 ^ @ -sin60 ^ @ cdot sin45 ^ @ + 1 #

# = (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) ^ 2) #

# = (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 #

Hy vọng nó giúp…

Cảm ơn bạn…

# tan105 ^ @ = tan (60 ^ @ + 45 ^ @) #

# => tan105 ^ @ = (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) #

# => (2tan52,5 ^ @) / (1-tan ^ 2 52,5 ^ @) = - 1 / (2-sqrt3) #

Để cho #tan52.5ucci@=x#

Hiện nay

# (2x) / (1-x ^ 2) = - 1 / (2-sqrt3) #

# => x ^ 2-2 (2-sqrt3) x-1 = 0 #

# => x = (2 (2-sqrt3) + sqrt (4 (2-sqrt3) ^ 2 + 4)) / 2 #

Như #52.5ucci@in "Rễ gốc Quadrant -ve bị bỏ qua" #

# => x = (2 (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 #