= Solution
Evenness means the negative half of the function is determined as well: this is the <triangular function> $q(x)=1-|x|$ on $|x|\leq1$, zero outside. Its transform is real and even:
$$
\widetilde q(k)=2\int_0^1(1-x)\cos(kx)\,dx.
$$
<Integration by parts>, or integrating the two terms separately, gives
$$
\boxed{\widetilde q(k)=\frac{2(1-\cos k)}{k^2}=\left(\frac{\sin(k/2)}{k/2}\right)^2,\qquad\widetilde q(0)=1.}
$$
The apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the <Fourier transform of a triangular function>.
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