Page 16 - 新一代技術高中數學C第三冊學習講義

Chapter 1 三角函數的應用 11

1-1 學習成效驗收 解析 P.3 ～ 5

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1. ண adb މቚԉdsin a = dcos b = dcos (a + b) ʘ࠽މ f
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2. ண a މୋɧ൥ࠢԉdb މୋ̬൥ࠢԉ˲ tan a = dsin b = - dۆ cot(a + b) ʘ࠽މ
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f

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3. Ӌ sin ( + i)cos (45c - i) + cos e + i osin (45c - i) ٙ࠽މ f
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4. νྡהͪdண aeb ѩމቚԉd˲ӊʃࣸѩމ͍˙Җdۆ
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a + b = f€౤ͪj΋Ӌ tan (a + b)

5. ʷᔊ cos 175ccos 25c + sin 175csin 25c = f

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6. ߰ 90c < i < 180c ˲ sin i = dۆ cos 2i ٙ࠽މ f
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7. ༊Ӌ sin 52.5c cos 52.5c cos 105c = f

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8. ߰ tan a + tan b = dtan (a + b) = 1dۆ tan a × tan b = f
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9. ʷᔊ tan 170c × tan 55c + tan 170c + tan 55c ʘ࠽މ f

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10. ண tan adtan b މ x - 3x + 2 = 0 ʘՇ࣬dۆ cos (a + b) ʘ࠽މ f

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11. ߰ f (x) = 2cos e - x o - 2cos x - 3 ٙ௰ɽ࠽݊ Mdۆ M = f
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12. ߰ aeb ޫމቚԉ˲ tan a = dcot b = 3d༊Ӌ sec (a + b) = f
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13. ண sin i - cos i = d༊Ӌ sin 2i = f
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14. ண tan 2i = dӋ tan i = f
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