How to derive the bank angle of an aircraft from its roll angle and pitch angle?

From Young (2017) (https://onlinelibrary.wiley.com/doi/book/10.1002/9781118534786) it is stated that we can define the bank angle ($Phi$) of an aircraft as the angle between its Y body axis and the horizontal plane. He then states the following equivalence:

$sin(Phi)=sin(phi)cos(theta)$

where $theta$ is the pitch angle of the aircraft and $phi$ is its roll angle. I want to derive this equivalence and this is my attempt so far which yields an alternative expression:

I define a global axis system $E=(O,X_e,Y_e,Z_e)$ where $O=begin{bmatrix} 0 0 0 end{bmatrix}$ is the origin of the system, and $X_e=begin{bmatrix} 1 0 0 end{bmatrix}$, $Y_e=begin{bmatrix} 0 1 0 end{bmatrix}$ and $Z_e=begin{bmatrix} 0 0 1 end{bmatrix}$ are orthonormal unit vectors which define North, East and ‘down’ respectively.

I also define an aircraft body axis system, $B=(O,X_b,Y_b,Z_b)$, whose starting orientation and position is coincident with $E$.

I first rotate $B$ about the $Y_b$ axis an angle $theta$, the pitch rotation, and then I rotate this rotated body axis system about its new $Xb$ axis an angle $phi$, the roll rotation. The relevant rotation matrices are:

$R_Y(theta)$=begin{bmatrix} cos(theta) & 0 & sin(theta) 0 & 1 & 0 -sin(theta) & 0 & cos(theta) end{bmatrix}

$R_X(phi)$=begin{bmatrix} 1 & 0 & 0 0 & cos(phi) & -sin(phi) 0 & sin(phi) & cos(phi) end{bmatrix}

which when applied in the correct order to specify the rotations outlined in the text above yields the composite matrix:

$R_Y(theta)R_X(phi)=R=$begin{bmatrix} cos(theta) & sin(theta)sin(phi) & sin(theta)cos(phi) 0 & cos(phi) & -sin(phi) -sin(theta) & cos(theta)sin(phi) & cos(theta)cos(phi) end{bmatrix}

I then have $B2=RB=R$ which is the body axis system after the rotations. $Y_{B2}= begin{bmatrix} sin(theta)sin(phi) cos(phi) cos(theta)sin(phi) end{bmatrix}$, the y axis of B2, and its projection onto the horizontal plane is $Y_{pB2}= begin{bmatrix} sin(theta)sin(phi) cos(phi) 0 end{bmatrix}$. I can then say that the cosine of the angle between them, the bank angle $Phi$, is the normalised dot product of the two vectors:

begin{align}
cos(Phi)=frac{Y_{B2}cdot Y_{pB2}}{|Y_{B2}||Y_{pB2}|}
end{align}

Doing the computation I am left with:

begin{align}
cos(Phi)=sin(theta)sin(phi) + cos(phi)
end{align}

Whereas I want:

begin{align}
sin(Phi)=sin(phi)cos(theta)
end{align}

Any advice on where I have gone wrong would be much appreciated.