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The exercise is 26E on Bartle´s Elements of Real Analysis.

It asks to use the fact that every continuous real valued function on $[0,\pi]$ is the uniform limit of a sequence of functions of the form: $$\sum_{k=1}^{m} a_kcos(kx)$$ To show that exery continuous real valued function $f$ on $[0,\pi]$ with $f(0)=f(\pi)$ is the uniform limit of a sequence of functions of the form: $$\sum_{i=1}^{n} b_ksin(kx)$$

The book also gives the next hint: If $f(0)=f(\pi)=0$, first approximate $f$ by a function $g$ vanishing on some intervals $[0,\sigma]$ and $[\pi-\sigma,\pi]$. Then consider $h(x)=\frac{g(x)}{sin x}$ for $x \in (0,\pi); h(x)=0$ for $x=0,\pi$

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Because $g(x)$ vanishes in intervals near $0$ and $\pi$, $h(x)$ will be continuous on the closed interval $[0,\pi]$ (show this) so that we may approximate $h(x)\approx \sum_{k=1}^ma_k\cos(kx)$. Then $g(x)\approx\sum_{k=1}^ma_k\sin(x)\cos(kx)$. Now, you should find a way of expressing $\sin(x)\cos(kx)$ only in terms of $\sin(ix)$ for different natural numbers $i$ (hint: it should end up being a sum of two different such terms). Of course, the approximations above must be made precise, but I leave that to you.

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  • $\begingroup$ How would you prove $ h(x)$ is continuous on $[0,\pi]$? That´s the only thing I´m missing. $\endgroup$ – PLanderos33 Oct 18 '19 at 14:58
  • $\begingroup$ For $x\in (0,\sigma)$, $g(x)=0$ so that $h(x)=0$. Thus, $h(x)$ will be continuous at $0$ by setting $h(0)=0$. $\endgroup$ – user293794 Oct 18 '19 at 18:31

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