幾題高中數學

(1) ∑n=1到無窮 (2/3)ncos(n120) = ?
(2/3)(-1/2) + (2/3)2(-1/2) + (2/3)3 + (2/3)4(-1/2) + (2/3)5(-1/2) + (2/3)6 + ...
= (-1/2)[(2/3) + (2/3)2 + (2/3)3 + ...] + (3/2)[(2/3)3 + (2/3)6 + (2/3)9 + ...]
= (-1/2)(2/3)/(1 - 2/3) + (3/2)(2/3)3/[1 - (2/3)3]
= -1 + (4/9)(27/19)
= -1 + 12/19
= -7/19
(2) ∑n=1到無窮 (1/2)nsin(n60)
設複數z = (1/2)(cos60 + i sin60) = 1/4 + √3/4i
zn = (1/2)n[cos(n60) + i sin(n60)]
∑n=1到無窮 (1/2)n[cos(n60) + i sin(n60)] = ∑n=1到無窮 zn
= z / (1 - z)
= (1/4 + √3/4i) / ( 3/4 - √3/4i)
= (1 + √3i) / ( 3 - √3i)
= (1 + √3i)(3 + √3i) / [(3 - √3i)(3 + √3i)]
= (3 + 3√3i + √3i - 3) / 12
= 4√3i/12
= √3i/3
∑n=1到無窮 (1/2)nsin(n60) = √3/3
(3) 三角形ABC中，若(a + 2b - 2c)2 + (a - 2b + c)2 = 0，則cosA:cosB:cosC=??
a + 2b - 2c = 0 ...(1)
a - 2b + c = 0 ...(2)
(1) + (2) => 2a - c = 0 => c = 2a
a - 2b + 2a = 0 => b = 3a/2
cosA = (9/4 + 4 - 1)/[(2)(3/2)(2)] = 7/8
cosB = (1 + 4 - 9/4)/[(2)(1)(2)] = 11/16
cosC = (1 + 9/4 - 4)/[(2)(1)(3/2)] = -1/4
cosA:cosB:cosC = 14/16 : 11/16 : -4/16 = 14 : 11 : - 4
(4) 三角形ABC中，三邊長為x^2+x+1，2x+1，x^2-1則最大角度為??
a = x2+x+1
b = 2x+1
c = x2-1
考慮cosA = (b2 + c2 - a2)/2bc
cosA = [(2x + 1)2 + (x2 - 1)2 - (x2 + x + 1)2] / [2(2x + 1)(x2 - 1)]
= [(2x + 1)2 + (x2 - 1 - x2 - x - 1)(x2 - 1 + x2 + x + 1)] / [2(2x + 1)(x2 - 1)]
= [(2x + 1)2 - (x + 2)(2x2 + x)] / [2(2x + 1)(x2 - 1)]
= [(2x + 1) - (x + 2)x] / [2(x2 - 1)]
= (1 - x2) / [2(x2 - 1)]
= -1/2
A = 120必然為最大角度
(5) 三角形ABC中，M為邊長BC之中點，若邊長AB=3，邊長AC=5，且角BAC=120度，試求tan角BAM=???
BC2 = AB2 + AC2 - 2(AB)(AC)cos120
= 9 + 25 - 30(-1/2) = 49
BC = 7
角B + 120 + 角C = 180 => 角C = 60 - 角B
AM / sinB = 3/sinAMB ... (1)
AM / sinC = 5/sinAMC
AM / sin(60 - B) = 5/ sin(180 - AMB) = 5 / sinAMB ... (2)
(1) / (2) => sin(60 - B)/sinB = 3/5
5sin60cosB - 5sinBcos60 = 3sinB
(5√3/2)cosB = 5.5sinB
tanB = (5√3/2)/5.5 => B = 38.21
AM2 = 32 + 3.52 - 2(3)(3.5)cos38.21 = 4.75
AM = 2.719
sinBAM/3.5 = sin38.21/2.719
sinBAM = (sin38.21/2.719)(3.5) = 0.7962
BAM = 52.8

(4).120度