急! 三角學13Q5

2012-06-22 1:48 am
請詳細步驟教我計以下幾條 :
(不要網址回答)


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回答 (1)

2012-06-22 7:47 am
✔ 最佳答案
15. (d)
tan240° + tan(-480°) - tan120°
= tan(180° + 60°) + tan(360° - 480°) - tan120°
= tan60° + tan(-120°) - tan120°
= tan60° - 2tan120°
= tan60° - 2tan(180 - 60)°
= 3tan60°
= 3√3


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16. (b)
tan95° tan105° tan115° tan125°tan135° tan145° tan155° tan165° tan175°
= tan(90+5)° tan(90+15)°..... tan(90+35)°tan(180-45)° tan(180-35)° ...... tan(180-5)°
= (-1/tan5) (-1/tan15°) (-1/tan25°) (-1/tan35°) tan(180-45)°(-tan35°) (-tan25°) (-tan15°) (-tan5°)
= -tan45°
= -1


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21.
tanθ sinθ - sinθ = 0
sinθ (tanθ - 1) = 0
sinθ = 0 or tanθ = 1
θ = 0°, 180°, 360° or θ =45°, (180+45)°
θ = 0°,45°, 180°, 225°, 360°


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22.
3 tanθ cosθ = -tanθ
3 tanθ cosθ + tanθ = 0
tanθ (3 cosθ + 1) = 0
tanθ = 0 or cosθ = -1/3
θ = 0°, 180°, 360° or θ = (180+70.53)°,(180-70.53)°
θ = 0°,109.47°, 180°, 250.53°, 360°


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23.
4 sinθ - 3 tanθ = 0
4 (sinθcosθ/cosθ) - 3 (sinθ/cosθ)= 0
(4 sinθ cosθ - 3 sinθ)/cosθ = 0
4 sinθ cosθ - 3 sinθ = 0
sinθ (4 cosθ - 3) = 0
sinθ = 0 or 4 cosθ = 3
sinθ = 0 or cosθ = 3/4
θ = 0°, 180°, 360° or θ = 41.41°, (360-41.41)°
θ = 0°, 41.41°, 180°, 318.59°, 360°


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24.
√2 sinθ = tanθ
√2 sinθ - tanθ = 0
(√2 sinθcosθ/cosθ) - (sinθ/cosθ) = 0
(√2 sinθcosθ - sinθ)/cosθ = 0
√2 sinθcosθ - sinθ = 0
sinθ (√2 cosθ - 1) = 0
sinθ = 0 or cosθ = 1/√2
θ = 0°, 180°, 360° or θ = 45°, (360-45)°
θ = 0°, 45°, 180°, 315°, 360°
參考: 胡雪


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