- 257 名前：１３２人目の素数さん [2024/03/12(火) 01:22:43.89 ID:pJAQFbPE.net]
- >>247
左辺は x = (x_1+x_2)/2 := m について 左右対称だから
a･cos(x) + b･sin(x) = r・cos(x-m),
と表わせる。ここに
r = ±√(aa+bb), cos(m) = a/r, sin(m) = b/r,
cos(x_1 + x_2) = cos(2m)
= 2{cos(m)}^2 - 1
= (aa-bb)/rr
= (aa-bb)/(aa+bb),
x_1 - m = m - x_2 := d, とおくと
cos(x_1 - x_2) = cos(2d)
= 2cos(d)^2 - 1
= 2(c/r)^2 - 1
= 2cc/(aa+bb) - 1,
