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[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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594: １３２人目の素数さん [sage] 2020/03/11(水) 09:12:19.97 ID:zi4olkqu

>>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^^
---------------------------------------------------------
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)},

http://rio2016.5ch.net/test/read.cgi/math/1581260776/594

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