I solved the relevant issues pertaining to this question - so this thread/question can be retired.
I am trying to derive the clockwise rotation linear combination...
So, the counter-clockwise rotation matrix is
\cos\alpha (-\sin\alpha)
\sin\alpha \cos\alpha
``derived" from \cos(\alpha + \beta), s.t.
x = r[\cos(\alpha + \beta)] = r[\cos\alpha \cos\beta - \sin\alpha \sin\beta]
= \cos\alpha (r \cos\beta) - \sin\alpha (r \sin\beta)
= X \cos\alpha - Y \sin\alpha.
Similarly,
y = r[\sin(\alpha + \beta) = r[\sin\alpha \cos\alpha + \cos\alpha \sin\beta]
= \sin\alpha (r \cos\beta) + \cos\alpha (r \sin\beta)
= X \sin\alpha + Y \cos\alpha.
Now, I know you make a linear combination out of these using matrix multiplication as I listed the counter-clockwise matrix above...and that you can just use the fact that a negative \beta will ``switch" the matrix to a clockwise rotation matrix, i.e., the same thing as multiplying the matrix by negative 1.
However, I am not able to solve correctly for y in the sine difference formula, if I start with the negative \beta in the first place, and this is where I need some help...
For starters, let's solve for x...as follows:
x = r[\cos(\alpha + (-\beta))] = r[\cos\alpha \cos(-\beta) - \sin\alpha \sin(-\beta)]
= r[\cos\alpha \cos\beta + \sin\alpha \sin\beta]
= \cos\alpha (r \cos\beta) + \sin\alpha (r \sin\beta)
= X \cos\alpha + Y \sin\alpha.
Now here's the rub, watch this...
y = r[\sin(\alpha + (-\beta))] = r[\sin\alpha \cos(-\beta) + \cos\alpha \sin(-\beta)]
= r[\sin\alpha \cos\beta - \cos\alpha \sin\beta]
= \sin\alpha (r \cos\beta) - \cos\alpha (r \sin\beta)
= X \sin\alpha - Y \cos\alpha.
As you see, this gives an incorrect clockwise rotation matrix, as follows:
\cos\alpha \sin\alpha
\sin\alpha (-\cos\alpha).
Whereas it should be...
\cos\alpha \sin\alpha
(-\sin\alpha) \cos\alpha.
Please advise, thanks,
carmamer