Derivation of equation (9.117) on Schutz's textbook

[MathematicalPhysicist]

On page 238 of his second edition of Schutz's he writes the following:

If we now add this to the incident wave, Eq. (9.107), we get the net result, to first order in $R$,
$$(9.117)\bar{h}^{net}_{xx}=\bar{h}_{xx}^{TT}+\delta \bar{h}^{TT}_{xx}=(A-2\pi \sigma m \Omega \ell_0 R \sin \phi)\cos [\Omega (z-t)-\psi]$$
where $$(9.118) \tan \psi = \frac{2\pi \sigma m \Omega \ell_0 R}{A} \cos \phi$$

Where Eq. (9.107) is: $$\bar{h}^{TT}_{xx}=A\cos (\Omega (z-t)) , \bar{h}_{yy}^{TT}=-\bar{h}^{TT}_{xx}$$
and $\delta \bar{h}^{TT}_{xx}=2\pi \sigma m \Omega \ell_0 R \sin [\Omega (z-t)-\phi]$.

Here's what I tried:
$$A\tan \psi / ( 2\pi \sigma m \Omega \ell_0 R )= \cos \phi $$
$$A\cos(\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R \sin ((\Omega (z-t)-\phi)=A\cos (\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R [\sin (\Omega (z-t))\cos \phi -\cos (\Omega (z-t))\sin \phi ] =$$
$$ (A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t)) + 2\pi \sigma m \Omega \ell_0 R \sin (\Omega (z-t)) \cos \phi = \frac{(A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t))\cos \psi + A \sin (\Omega (z-t)) \sin \psi}{\cos \psi}$$

How to proceed to get the above identity in equation (9.117), I don't see it.
Can you help me?

Thanks in advance!

 

[David_Zheng]

For simplifying terms, following algebra is applied:
let:$$A=A$$ $$B=2\pi\sigma m\Omega\ell_{0}R$$and$$\Omega\left(z-t\right)=\tau$$
Equations you posted can be wirtten as:
$$\bar{h}_{xx}^{TT}=A \cos\tau$$
$$\delta\bar{h}_{xx}^{TT}=B \sin \left(\tau-\phi\right)$$
Then:
$$\bar{h}_{xx}^{TT}+\delta\bar{h}_{xx}^{TT}=A \cos\tau+B \sin \left(\tau-\phi\right)$$
$$=A \cos\tau+B\cos\phi\sin \tau-B\sin\phi\cos\tau$$
$$=\left(A-B\sin\phi\right) \cos\tau+B\cos\phi\sin \tau...(Eq1)$$
Eqs.(9.117) and Eqs.(9.118) are:
$$\bar{h}_{xx}^{TT}+\delta\bar{h}_{xx}^{TT}=\left(A-B\sin\phi\right)\cos\left(\tau-\psi\right)...(Eq2)$$
$$\tan\psi=\frac{B\cos\phi}{A}...(Eq3)$$
Expand Eq2 and combining Eq3:
$$\bar{h}_{xx}^{TT}+\delta\bar{h}_{xx}^{TT}=\left(A-B\sin\phi\right)\cdot\left(\frac{A}{r}\cdot\cos\tau+\frac{B\cos\phi}{r}\cdot\sin\tau\right)...(Eq4)$$
where:$$r=\left(A^{2}+\left(B\cos\phi\right)^{2}\right)^{\frac{1}{2}}$$
Comparing Eq1&Eq4, coefficients of term $\cos\tau$ and $\sin\tau$ must be balanced, if the book is right or some relationship between A and B exsit, following equation must be valid:
$$\left(A-B\sin\phi\right)=\left(A-B\sin\phi\right)\cdot\frac{A}{r}...Cond.(1)$$
$$B\cos\phi=\left(A-B\sin\phi\right)\cdot\frac{B\cos\phi}{r}...Cond.(2)$$
To my knowledge, the above equations are weird, on the other hand, can you tell more details of A and B?

 

[MathematicalPhysicist]

@David_Zheng , well, more details that I see that are relevant are $R=\mathcal{O}(h_{xx}^{TT}\ell_0)$, $\sigma$ is proportional to $\exp(-\epsilon r)$.
$A$ is the amplitude where two masses oscillate , $A$ is the amplitude that each mass in the $\sigma$ masses moves, one-half of the the total stretching of the spring $R$.

Do you want me to put on the screen scans of this part of the book?