Compensate for FIR filter attenuation
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:

$$ y[n] = M \sum_{k=0}^{N} { b_k \cos( \alpha (n-k) + \phi ) }  $$

$$ y[n] = M \sum_{k=0}^{N} { b_k [  \cos( \alpha n + \phi ) \cos( \alpha k ) + \sin( \alpha n + \phi ) \sin( \alpha k ) ] }  $$

$$ y[n] = M \cos( \alpha n + \phi ) \sum_{k=0}^{N} { b_k  \cos( \alpha k ) } + M \sin( \alpha n + \phi ) \sum_{k=0}^{N} { b_k \sin( \alpha k ) }  $$

$$ y[n] = A \cos( \alpha n + \phi ) + B  \sin( \alpha n + \phi )  $$

$$ A = M  \sum_{k=0}^{N} { b_k  \cos( \alpha k ) } $$

$$ B = M  \sum_{k=0}^{N} { b_k  \sin( \alpha k ) } $$

$$ y[n] = M_2 \cos( \alpha n + \phi + \theta ) $$

$$ y[n] = M_2 \cos( \theta ) \cos( \alpha n + \phi ) - M_2 \sin( \theta ) \sin( \alpha n  + \phi ) $$

$$ A = M_2 \cos( \theta ) $$

$$ B = -M_2 \sin( \theta ) $$

$$ A^2 + B^2 = M_2^2 = M^2 \left[ \left( \sum_{k=0}^{N} { b_k \cos( \alpha k ) } \right)^2 + \left( \sum_{k=0}^{N} { b_k \sin( \alpha k ) } \right)^2 \right] $$

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