SOLUTION: please please help :define a function f:[0,2pi]->real number such that f(0)=f(2pi).show that there exist an element x in [0,pi] such that f(x)=f(x+pi). thanks a lot.
Algebra.Com
Question 811967: please please help :define a function f:[0,2pi]->real number such that f(0)=f(2pi).show that there exist an element x in [0,pi] such that f(x)=f(x+pi). thanks a lot.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Define \ =\ f(2\pi))
Let \ =\ \sin(x))
\left(y\ =\ \sin(x)\ \rightarrow \ y\ \in\ \[-1,1\]\ \in\ \mathbb{R}\right))
and
\ =\ 0\ \small{\wedge}\ \Large \sin(2\pi)\ =\ 0)
Show \ =\ f(x\ +\ \pi))
Let 
Then \ =\ 0\ \small{\wedge}\ \Large f(\pi)\ =\ 0\ \small{\wedge}\ \Large f(2\pi)\ =\ 0)
Interestingly enough, the above result notwithstanding:
\ \left(\sin(x)\ =\ -\sin(x\ +\ \pi)\right))
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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