how to convince myself cos(A-B) proof works for all positions of A and B?

Question

I managed to work this proof using a diagram. Then when I googled I found similar proofs. But aren't they incomplete? Because they prove the formula works only in first quadrant. Why do textbooks claim these are valid proofs even though they work only for a particular arrangement of rays $a$ and $b$ ?
How do we know that one proof is sufficient for all arrangements of rays $a$ and $b$ ?

J.G. · Answer

You can use the rotation matrix$$left(begin{array}{cc}
cosleft(A+Bright) & -sinleft(A+Bright)\
sinleft(A+Bright) & cosleft(A+Bright)
end{array}right)=left(begin{array}{cc}
cos A & -sin A\
sin A & cos A
end{array}right)left(begin{array}{cc}
cos B & -sin B\
sin B & cos B
end{array}right)$$if you first prove an anticlockwise rotation by $A$ radians moves $left(begin{array}{c}
x\
y
end{array}right)$ to $left(begin{array}{cc}
cos A & -sin A\
sin A & cos A
end{array}right)left(begin{array}{c}
x\
y
end{array}right)$. But this is trivial from the circle-based definitions of the trigonometric functions: just separately check the cases $x=1,,y=0$ and $x=0,,y=-1$ (obtained from the first case with a anticlockwise right angle), and use linearity. Note if we compare this reasoning with $e^{i(A+B)}=e^{iA}e^{iB}$, we can read off a matrix representation of complex numbers:$$1=left(begin{array}{cc}
1 & 0\
0 & 1
end{array}right),,i=left(begin{array}{cc}
0 & -1\
1 & 0
end{array}right).$$

how to convince myself cos(A-B) proof works for all positions of A and B?

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