[過去ﾛｸﾞ] 分からない問題はここに書いてね458 (1002ﾚｽ)
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586(3): １３２人目の素数さん [sage] 2020/03/11(水) 02:45:12.90 ID:y9Jt3QH1(1/5) AAS
>>579 , >>581
これどうやって示すのか誰か教えてください.
x+y+z=π, tan(x) + tan(y) + tan(z) = tan(x)tan(y)tan(z)
a = cot(x), A = ... =1/sin(x) + 1 - cot(x), B= . . .
これで行くのかなと予想は立てたもののスマートな式変形が思い浮かびません.
591(1): １３２人目の素数さん [] 2020/03/11(水) 08:14:52.76 ID:avK6eeO9(1) AAS
>>586
>x+y+z=π, tan(x) + tan(y) + tan(z) = tan(x)tan(y)tan(z)
tan(x+y+z)=0
(tan x+tan y+tan z-tan x tan y tan z)/(1-tan x tan y-tan y tan z-tan z tan x)=0
tan x+tan y+tan z-tan x tan y tan z=0
594(1): １３２人目の素数さん [sage] 2020/03/11(水) 09:12:19.97 ID:zi4olkqu(1/2) AAS
>>586
a = cot(x), b = cot(y), c = cot(z),
とおけば
1/a + 1/b + 1/c - 1/abc = tan(x)+tan(y)+tan(z) - tan(x)tan(y)tan(z)
= sin(x+y+z)/{cos(x)cos(y)cos(z)},
A-1 = tan(x/2), B-1 = tan(y/2), C-1 = tan(z/2),
より
(与式) = (A-1)(B-1) + (B-1)(C-1) + (C-1)(A-1) - 1
= tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) - 1
= -cos((x+y+z)/2)/{cos(x/2)cos(y/2)cos(z/2)}
= -(1/a +1/b +1/c -1/abc)･cos(x)cos(y)cos(z)/{2sin((x+y+z)/2)cos(x/2)cos(y/2)cos(z/2)}
ですね^^
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加法公式の略証
e^{i(x+y+z)} = e^(ix)･e^(iy)･e^(iz)
= {cos(x)+i･sin(x)}{cos(y)+i･sin(y)}{cos(z)+i･sin(z)}
= cos(x)cos(y)cos(z){1+i･tan(x)}{1+i･tan(y){1+i･tan(z)},
実部から
cos(x+y+z) = cos(x)cos(y)cos(z) - sin(x)sin(y)cos(z) - cos(x)sin(y)sin(z) - sin(x)cos(y)sin(z)
= cos(x)cos(y)cos(z){1 - tan(x)tan(y) - tan(y)tan(z) - tan(z)tan(x)},
虚部から
sin(x+y+z) = sin(x)cos(y)cos(z) + cos(x)sin(y)cos(z) + cos(x)cos(y)sin(z) - sin(x)sin(y)sin(z)
= cos(x)cos(y)cos(z){tan(x) + tan(y) + tan(z) - tan(x)tan(y)tan(z)},
595: 586 [sage] 2020/03/11(水) 09:20:39.62 ID:y9Jt3QH1(2/5) AAS
tanの加法を二回使って
tan(x+y+z) = (tx + ty + tz - txtytz) / (1 - txty - tytz - tztx)
[公式1] x+y+z = π → tanx + tany + tanz = tanx tany tanz
[公式2] x+y+z = π/2 → tanx tany + tany tanz + tanz tanx = 1
a = cot(x) (0<x<π/2), b= ... と置ける. {∵ 公式1}
A = 1 + 1/sin(x) - cot(x) = 1 + X,
X := (1-cos(x))/sin(x) = 2 sin(x/2)^2 / sin(x) = sin(x/2)/cos(x/2) = tan(x/2)
(2-A)(2-B)(2-C) + ABC = {2(AB+BC+CA) - 4(A+B+C-1)} + 4
(2-A)(2-B)(2-C) + ABC = (1-X)(1-Y)(1-Z) + (1+X)(1+Y)(1+Z)
= 2 + 2(XY+YZ+ZX) = 2 + 2*1 = 4 {∵ 公式2}
∴ (AB+BC+CA) - 2(A+B+C-1) = {(2-A)(2-B)(2-C) + ABC}/2 -2 = 2 - 2 = 0
できた. 幾何学的意味などがあるなら教えて欲しい.
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