標題:
急! 三角學13Q5
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最佳解答:
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 ===== 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 ===== 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° ===== 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° ===== 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° ===== 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°
其他解答:7638E7481407D16B
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