general solution一問

用戶117714

2007-03-08 5:15 am

在general solution,用不同trigo function 所solve的ans 都有所不同,
可是如何先能核算ans?? e.g.
2(sina+cosa)=√ 6
以sin functon,
4(1-2cosasina)=6
sin2a=1/2
therefore,we get a=(nπ)/2+(-1)^n(1/4)

but the ans is 2nπ+π/12 and 2nπ+5π/12(by cos)
how can I porve that my sin function ans?

回答 (1)

chaoseng2000

2007-03-08 5:21 pm

✔ 最佳答案

by sin
2(sina+cosa)=√ 6
以sin functon,
4(1+2cosasina)=6-----------錯了
sin2a=1/2 -----------(1)
therefore,we get 2a=nπ+(-1)^n(π/6)
a=(nπ)/2+(-1)^n(π/12)---------(2)-----------錯了

by cos
(1)^2
(sin2a)^2=1/4
1-(cos2a)^2=1/4
cos2a=±√3/2
therefore,we get 2a=2nπ+π/6,2a=2nπ+5π/6
a=nπ+π/12,a=nπ+5π/12<-----------錯了

當n為奇數時,設n=2k-1(k為整數)
(2)=(2k-1)π/2-(π/12)
=2kπ/2-π/2-π/12
=kπ-7π/12
=(k-1)π+5π/12--------(因k-1為整數)
即與nπ+5π/12相同

當n為偶數時,設n=2k(k為整數)
(2)=2kπ/2+(π/12)
=kπ+π/12--------------(因k為整數)
即與nπ+π/12相同

所以無論用什麼方法,方程的解必相同

希望能幫到你!

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