Решите систему: x+y=3pi/4
tg x - tg y =2
По формуле разности тангенсов:
sin(x-y) / (cosx*cosy) = 2
sin(x-y) = 2cosx*cosy
sin(x-y) = cos(x+y) + cos(x-y)
И с учетом x+y=3pi/4:
sin(2x-3pi/4) - cos(2x - 3pi/4) = cos(3pi/4)
Теперь воспользуемся формулой перехода:
Asina + Bcosa = кор(A^2+B^2)*sin(a+ arctg(B/A)):
(кор2)sin(2x-3pi/4 - pi/4) = -(кор2)/2
sin(pi - 2x) = 1/2
sin2x = 1/2
2x = (-1)^k *pi/6 + pik
x = (-1)^k *pi/12 + pik/2, y = 3pi/4 - (-1)^k *pi/12 - pik, k: Z
x + y = 3*π/4
tg x - tg y = 2
tg x - tg(3*π/4 - x) = 2
tg x - (tg 3*π/4 - tg x)/(1 + tg 3*π/4 * tg x) = 2
tg x - (-1 - tg x) / (1 - tg x) - 2 = 0
tg x + (1 + tg x)/(1 - tg x) - 2 = 0
Если tg x ≠ 1 , то
tg x * (1 - tg x) + 1 + tg x - 2 * (1 - tg x) = 0
tg x - tg²x + 1 + tg x - 2 + 2 * tg x = 0
-tg²x + 4 * tg x - 1 = 0
tg²x - 4 * tg x + 1 = 0
tg x = 2 ± √ 3
x = arctg (2 ± √ 3) + π * n
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Verified answer
По формуле разности тангенсов:
sin(x-y) / (cosx*cosy) = 2
sin(x-y) = 2cosx*cosy
sin(x-y) = cos(x+y) + cos(x-y)
И с учетом x+y=3pi/4:
sin(2x-3pi/4) - cos(2x - 3pi/4) = cos(3pi/4)
Теперь воспользуемся формулой перехода:
Asina + Bcosa = кор(A^2+B^2)*sin(a+ arctg(B/A)):
(кор2)sin(2x-3pi/4 - pi/4) = -(кор2)/2
sin(pi - 2x) = 1/2
sin2x = 1/2
2x = (-1)^k *pi/6 + pik
x = (-1)^k *pi/12 + pik/2, y = 3pi/4 - (-1)^k *pi/12 - pik, k: Z
Verified answer
x + y = 3*π/4
tg x - tg y = 2
x + y = 3*π/4
tg x - tg(3*π/4 - x) = 2
tg x - (tg 3*π/4 - tg x)/(1 + tg 3*π/4 * tg x) = 2
tg x - (-1 - tg x) / (1 - tg x) - 2 = 0
tg x + (1 + tg x)/(1 - tg x) - 2 = 0
Если tg x ≠ 1 , то
tg x * (1 - tg x) + 1 + tg x - 2 * (1 - tg x) = 0
tg x - tg²x + 1 + tg x - 2 + 2 * tg x = 0
-tg²x + 4 * tg x - 1 = 0
tg²x - 4 * tg x + 1 = 0
tg x = 2 ± √ 3
x = arctg (2 ± √ 3) + π * n