= Solution
The input is even, so its <Fourier transform> is real and even. Decomposing the cosine into <complex exponentials> and integrating over its support gives
$$
\widetilde f(k)=\frac{\sin[(k-1)a]}{k-1}+\frac{\sin[(k+1)a]}{k+1}.
$$
Combining the fractions and using the angle addition formula yields
$$
\boxed{\widetilde f(k)=\frac{2k\cos a\sin(ak)-2\sin a\cos(ak)}{k^2-1},\qquad\lambda=2,\quad\mu=-2.}
$$
The apparent <poles> at $k=\pm1$ are <removable singularities>; either the integral or the sine-quotient limit gives $\widetilde f(\pm1)=a+\sin(2a)/2$.
If $\cos a\ne0$, expansion of the denominator gives
$$
\widetilde f(k)=\frac{2\cos a\sin(ak)}k-\frac{2\sin a\cos(ak)}{k^2}+O(|k|^{-3}).
$$
The oscillatory envelope is therefore $O(|k|^{-1})$. At $a=\pi/2+n\pi$, $n=0,1,2,\ldots$, the endpoint values vanish and the first term cancels:
$$
\boxed{\cos a=0:\quad\widetilde f(k)=-\frac{2\sin a\cos(ak)}{k^2}+O(|k|^{-4}).}
$$
This gives an $O(|k|^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first <derivative>. Each additional degree of endpoint smoothness gives another inverse power, as explained by <Fourier decay from a derivative jump>. The exact example is the <Fourier transform of a truncated cosine>; isolated zeros of its oscillatory factors do not change the envelope order.
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