標題:
發問:
3cosθ + 4sinθ = 2 0°≦θ≦360°
最佳解答:
As follows~~~ As follows~~~ For 0 <= θ <= 360 Let rcos(θ - α) =3cosθ + 4sinθ (rcosα)cosθ + (rsinα)sinθ= 3cosθ + 4sinθ So, rcosα = 3 ─── (1) rsinα = 4 ─── (2) (2) / (1): tanα = 4 / 3 α = 53.13 r = 5 Therefore, 3cosθ + 4sinθ= 2 5cos(θ – 53.13) = 2 cos(θ – 53.13) = 2 / 5 θ – 53.13 = 360n +-66.42 θ = 360n + 119.55 or360n – 13.29 (where n is an integer) (cor. to 2 d.p.)
其他解答:
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唔識做a.maths發問:
3cosθ + 4sinθ = 2 0°≦θ≦360°
最佳解答:
As follows~~~ As follows~~~ For 0 <= θ <= 360 Let rcos(θ - α) =3cosθ + 4sinθ (rcosα)cosθ + (rsinα)sinθ= 3cosθ + 4sinθ So, rcosα = 3 ─── (1) rsinα = 4 ─── (2) (2) / (1): tanα = 4 / 3 α = 53.13 r = 5 Therefore, 3cosθ + 4sinθ= 2 5cos(θ – 53.13) = 2 cos(θ – 53.13) = 2 / 5 θ – 53.13 = 360n +-66.42 θ = 360n + 119.55 or360n – 13.29 (where n is an integer) (cor. to 2 d.p.)
其他解答:
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