Anonymous 寫到:Let s(x) and c(x) be two functions satisfying s’(x)=c(x) and
c’(x)= -s(x) for all x. If s(0)=0 and c(0)=1, prove that
[s(x)]^2 + [c(x)]^2 = 1.
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Re: [問題]請問如何解
由 訪客 於 星期五 十二月 16, 2005 11:54 am
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由 一陣風 於 星期五 十二月 16, 2005 12:18 am
Prove that [s(x)]^2 + [c(x)]^2 is a constant function first.
Let F(x) = [s(x)]^2 + [c(x)]^2, then
F'(x) = 2s(x)s'(x) + 2c(x)c'(x) =2s(x)c(x) - 2c(x)s(x) = 0
Hence F(x) is a constant.
Since s(0) = 0 and c(0) = 1, F(x) = F(0) = 0^2 + 1^2 = 1 for all x.
Let F(x) = [s(x)]^2 + [c(x)]^2, then
F'(x) = 2s(x)s'(x) + 2c(x)c'(x) =2s(x)c(x) - 2c(x)s(x) = 0
Hence F(x) is a constant.
Since s(0) = 0 and c(0) = 1, F(x) = F(0) = 0^2 + 1^2 = 1 for all x.
[問題]請問如何解
由 訪客 於 星期四 十二月 15, 2005 11:17 pm
Let s(x) and c(x) be two functions satisfying s’(x)=c(x) and
c’(x)= -s(x) for all x. If s(0)=0 and c(0)=1, prove that
[s(x)]^2 + [c(x)]^2 = 1.
c’(x)= -s(x) for all x. If s(0)=0 and c(0)=1, prove that
[s(x)]^2 + [c(x)]^2 = 1.