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Sa se determine functiile continue \( f:[-1,1]\to\mathbb{R} \) cu proprietatea \( \int_0^nf(\cos t)dt=f(\sin n) \) pentru orice \( n\in\mathbb{N} \).

GM 12/1999

The sequence \( (\sin n) \) is dense in \( [-1;1] \).

For any \( x \in [-1;1] \) there exist a sequence \( (n_{k}) \) such that
the sequence \( (sin n_{k}) \) converges to \( x \).

\( \int_{0}^{n_{k}}f(\cos t)dt=f(\sin n_{k}) \)

\( f(x)=\int_{0}^{\infty}f(\cos t)dt \) for any \( x \in [1;-1] \)

f is constant C

nC=C for any n in N

f=0

What about \( f(x) = x \) ?

\( \sin n=\sin(n+2p\pi) \) for any \( p \in \mathbb{Z} \).

So we can extend the given formulae to \( \mathbb{N}+2 \pi \mathbb{Z} \).

For any \( n \in \mathbb{N},\ p \in \mathbb{Z} \) we have \( \int_{0}^{n+2p \pi}f(\cos t)dt=f(\sin(n+2p \pi)) \).

\( N+2 \pi Z \) is dense in R, for any x in R there exist \( a_{k} \) sequence of N, there exist \( b_{q} \) in Z such that \( a_{k}+2b_{q}\pi \) converge to x

Then for any x in R, \( \int_{0}^{x}f(\cos t)dt=f(\sin x) \) (1)

(1) imply f is odd function

LHS of (1) is derivable so RHS also

\( \cos x f^{\prime}(\sin x)=f(\cos x) \)(2)

Change x in \( x-\frac{\pi}{2} \) gives
\( \sin x f^{\prime}(-\cos x)=f(\sin x)=\sin x f^{\prime}(\cos x) \)(3) since f' is even function

Let \( h(x)=f(\sin x),g(x)=f(\cos x) \) are derivable

g'=-h
h'=g

we get

g''+g=0

\( g(x)=A\cos x +B\sin x \) with g even function B=0

\( g(x)=A\cos x =f(\cos x) \)

\( f(t)=At \) with A any real number