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Determine the nodes of the first degree interpolating polynomial which approximate $f(x) = cos(x)$ the best on $[-\pi,\pi]$ in the uniform norm. What is the interpolation error?

I think I need a polynomial such that $$p(-\pi)=-1, p(0)=1 $$ Further I get $$p[-\pi,0]=(1+1)/\pi=2/\pi$$ and $$p(x) = -1 +(x+\pi)2/\pi$$ So my nodes are $-\pi$ and $0$. Is that correct or am I wrong ? The interpolation error is $$sin(z)(x+\pi)(x-\pi)x/6 \le (x^2-\pi^3)x/6 \le \pi^2/16 $$ with $x\in[-\pi,0]$ (where i chose $x=-\pi/2$).

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    It seems that you found some first trial function for the interval $[-π,0]$ while the task is asking for a solution on $[-π,π]$. – 2017-02-17
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    The problem is that if I use $p(\pi)=-1$ and get $p[-\pi,0,\pi]$ then p would be a second degree polynomial and not a first degree polynomial – 2017-02-17
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    Compare your result to the zero polynomial. Look up the [Chebyshev alternation theorem](https://www.encyclopediaofmath.org/index.php/Chebyshev_theorem) and the [Remez exchange algorithm](https://en.wikipedia.org/wiki/Remez_algorithm). – 2017-02-17

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