化下列名三角內函數為銳角三角函數:
1. cos ( 7兀/5)
2. csc (- 9兀/5)
3. tan (8兀/3)
4. sec (- 9兀/4)
F.4 附加數學 - 三角函數(高手請進)
2007-11-20 8:42 am
回答 (2)
2007-11-20 10:03 am
✔ 最佳答案
1.cos (7π / 5)
= cos (π + 2π / 5)
= -cos (2π / 5)
2.
csc (-9π / 5)
= -csc (9π / 5)
= -csc (π + 4π / 5)
= csc (4π / 5)
= csc (π - π / 5)
= csc (π / 5)
3.
tan (8π / 3)
= tan (2π + 2π / 3)
= tan (2π / 3)
= tan (π - π / 3)
= -tan (π / 3)
4.
sec (-9π / 4)
= sec (9π / 4)
= sec (2π + π / 4)
= sec (π / 4)
有錯請指正!
參考: My Maths knowledge
2007-11-20 7:19 pm
1. cos ( 7兀/5)
= cos (兀 + 2兀/5)
= - cos (2兀/5)
2. csc (- 9兀/5)
= csc (-2兀 + 兀/5)
= csc (兀/5)
3. tan (8兀/3)
= tan (2兀 + 2兀/3)
= tan (2兀/3)
= tan (兀 - 兀/3)
= - tan (兀/3)
4. sec (- 9兀/4)
= sec (-2兀 - 兀/4)
= sec (-兀/4)
= sec (兀/4)
= cos (兀 + 2兀/5)
= - cos (2兀/5)
2. csc (- 9兀/5)
= csc (-2兀 + 兀/5)
= csc (兀/5)
3. tan (8兀/3)
= tan (2兀 + 2兀/3)
= tan (2兀/3)
= tan (兀 - 兀/3)
= - tan (兀/3)
4. sec (- 9兀/4)
= sec (-2兀 - 兀/4)
= sec (-兀/4)
= sec (兀/4)
收錄日期: 2021-04-13 14:32:19
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