FIR definition:
$$ y[n] = \sum_{k=0}^{N} { b_k x[n-k] } $$
Sinusoid signal definition:
$$ x[n] = M \cos( \alpha n + \phi ) $$
A whole bunch of math:
$$ y[n] = M \sum_{k=0}^{N} { b_k \cos( \alpha (n-k) + \phi ) } $$
$$ y[n] = M \sum_{k=0}^{N} { b_k [ \cos( \alpha n + \phi ) \cos( \alpha k ) - \sin( \alpha n + \phi ) \sin( \alpha k ) ] } $$$$ y[n] = M \sum_{k=0}^{N} { b_k [ \cos( \alpha n + \phi ) \cos( \alpha k ) + \sin( \alpha n + \phi ) \sin( \alpha k ) ] } $$
$$ y[n] = M \cos( \alpha n + \phi ) \sum_{k=0}^{N} { b_k \cos( \alpha k ) } - M \sin( \alpha n + \phi ) \sum_{k=0}^{N} { b_k \sin( \alpha k ) } $$$$ y[n] = M \cos( \alpha n + \phi ) \sum_{k=0}^{N} { b_k \cos( \alpha k ) } + M \sin( \alpha n + \phi ) \sum_{k=0}^{N} { b_k \sin( \alpha k ) } $$
$$ y[n] = A \cos( \alpha n + \phi ) - B \sin( \alpha n + \phi ) $$$$ y[n] = A \cos( \alpha n + \phi ) + B \sin( \alpha n + \phi ) $$
$$ A = M \sum_{k=0}^{N} { b_k \cos( \alpha k ) } $$
$$ B = M \sum_{k=0}^{N} { b_k \sin( \alpha k ) } $$
$$ y[n] = M_2 \cos( \alpha n + \phi + \theta ) $$
$$ y[n] = M_2 \cos( \theta ) \cos( \alpha n + \phi ) - M_2 \sin( \theta ) \sin( \alpha n + \phi ) $$
$$ A = M_2 \cos( \theta ) $$
$$ B = M_2 \sin( \theta ) $$$$ B = -M_2 \sin( \theta ) $$
$$ A^2 + B^2 = M_2^2 = M^2 \left[ \left( \sum_{k=0}^{N} { b_k \cos( \alpha k ) } \right)^2 + \left( \sum_{k=0}^{N} { b_k \sin( \alpha k ) } \right)^2 \right] $$
Your desired equation:
$$ \frac{M}{M_2} = \frac{ 1 }{ \sqrt{ \left( \sum_{k=0}^{N} { b_k \cos( \alpha k ) } \right)^2 + \left( \sum_{k=0}^{N} { b_k \sin( \alpha k ) } \right)^2 } } $$
I think I've done the math right. The $b_k$s are your FIR coefficients and $\alpha$ is your frequency in radians per sample. $M_2$ is the amplitude of your smoothed sinusoid and $M$ is the original amplitude.
So you want multiply your smoothed results by $ \frac{M}{M_2} $.
Hope this helps. I just did this and haven't tested it.
Ced