F.4 A-MATHS,,

2007-02-26 12:16 am

利用2sinacosb=sin(a+b)+sin(a-b)

證明2(sin@-sin3@+sin5@-sin7@+sin9@)cos@=sin10@,

由此,解方程sin@+sin5@+sin9@=sin3@+sin7@

其中0<@<90

回答 (1)

Chiu

2007-02-26 4:27 am

✔ 最佳答案

(a) Prove 2(sin@-sin3@+sin5@-sin7@+sin9@)cos@=sin10@
Proof: 2(sin@-sin3@+sin5@-sin7@+sin9@)cos@
= 2sin@cos@ - 2sin3@cos@ + 2sin5@cos@ - 2sin7@cos@ +2sin9@cos@
= [sin(@+@)+sin(@-@)] - [sin(3@+@)+sin(3@-@)] + [sin(5@+@)+sin(5@-@)]
- [sin(7@+@)+sin(7@-@)] + [sin(9@+@)+sin(9@-@)]
= (sin2@ + sin 0) - (sin4@ + sin2@) + (sin6@ + sin4@) - (sin8@ + sin6@) +
(sin10@ - sin 8@)
= sin 10@

(b) sin@+sin5@+sin9@=sin3@+sin7@
=> sin@+3@+sin5@-sin7@+sin9@=0
=> 2cos@(sin@+3@+sin5@-sin7@+sin9@)=0 x 2cos@
=> sin 10@ = 0 (by (a))
=> 10@ = 180n where n = ..., -1, 0, 1, 2, ...
=> @ = 18n where n = 1, 2, 3, 4 since 0 < @ < 90

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