數學問題(三角函數)

Question

已知tanx+cotx=4 且0度 (算法)~^^

Answer 1

tanx + cotx = ( sinx / cosx ) + ( cosx / sinx ) = ( sin^2 x + cos^2 x ) / sinx*cosx = 4

→ 1 / (2*sinx*cosx) = 2 → 1 / sin2x = 2 → sin2x = 1/2 → 2x = 30度

→ x = 15度

2006-10-08 22:34:50 補充：
(1) sin^2 x + cos^2 x = 1(2) 2*sinx*cosx = sin2x(3) sin2x = 1/2 → 2x = sin^-1 (1/2) = 30度

2006-10-09 10:56:01 補充：
x = 15度 → sinx = 1 / (8+4根號3) cosx = (2+根號3) / (8+4根號3)tanx = sinx / cosx = 1 / (2+根號3) = 2 - 根號3

Answer 2

tanθ+cotθ= sinθ/ cosθ+ cosθ/ sinθ= sinθ平方+ cosθ平方 / sinθcosθ
= 1 / sinθcosθ→ sinθcosθ= 1 / 4

(sinθ+ cosθ) 平方 = 1+ 2 sinθcosθ = 3 / 2 → sinθ+ cosθ= ± 3 / 2 (負不和 因為0度<θ<45度) — "1"

(sinθ－ cosθ) 平方 = 1－2 sinθcosθ=1 / 2 → sinθ－ cosθ= ± 1 / 2 (正不和 因為0度<θ<45度)→ "2"

由 "1" "2" 知 sinθ = 1 / 2 , cosθ=1 → tanθ= sinθ / cosθ= 1 / 2


不知道有沒有算對 希望能幫上忙