F.4 附加數學 - 三角方程之通解

1) tan 5θ = cot 4θ
cos4θ cos5θ﹣sin4θ sin5θ = 0
cos(4θ + 5θ) = 0
cos 9θ = 0
9θ = 360n°± 90°
θ = 40n°± 10°
或相等地，
θ = 20n°± 10°

2) sin θ + cosθ = √2
√2 cos(θ﹣α) = √2
α = tan-1(√2/√2)
α = 45°
√2 cos(θ﹣45°) = √2
cos(θ﹣45°) = 1
θ﹣45°= 360n°
∴θ = 360n°+ 45°

3)cos x﹣√3 sinx = 1
√(12 + 3)cos(x﹣α) = 1
2cos(x﹣α) = 1
α = tan-1(-√3/1)
α = -60°
2cos(x + 60°) = 1
x + 60°= 360n°± 60°
x = 360n°或 x = 360n°﹣120°
或相等地，
x = 360n°,360n°+240°

4)√3 sin θ﹣cosθ = √2
√(3+1) cos(θ﹣α) =-√2
2cos(θ﹣α) = 3
α = tan-1(-√3)
α = - 60°
2cos(θ + 60°) = -√2/2
θ + 60° = 360n°±135°
∴θ = 360n°﹣195°或 360n°+ 75°
θ = 360n°+ 165°或 360n°+ 75°

5) 4sin θ + 5cosθ = 3
√(42 +52 )cos(θ﹣α) = 3
√41 cos(θ﹣α) = 3
α = tan-1(4/5)
α = 38.66°
√41 cos(θ﹣38.66°) = 3
cos(θ﹣38.66°) = 3 / √41
θ﹣38.66° = 360n°± 62.06°
∴θ = 360n°+ 100.7°或 360n°﹣23.4°