I've a few questions that stem from the proof given in my textbook regarding the cosine difference identity.
The proof goes like this:
Let $\alpha$ and $\beta$ be angles plotted in standard position on the Unit Circle, where $\alpha \geqslant\beta $. (Ignore the $\alpha _0$ and $\beta _0$ in the diagram, just take them to mean $\alpha$ and $\beta$). Let $P$ be the point on the terminal side of $\alpha$ that lies on the Unit Circle, and $Q$ be the point the terminal side of $\beta$ that lies on the Unit Circle.
This gives:
Plotting the angle $\alpha -\beta$ in standard position gives:
Let $A$ be the point on the terminal side of $\alpha -\beta$ that lies on the Unit Circle, and $B$ be the point where the initial side of the angle $\alpha -\beta$ intersects the Unit Circle.
As the angles $POQ$ and $AOB$ are congruent this must mean the chords $PQ$ and $AB$ are also equal, and it is possible to set up an equivalance relationship between the two. The distance formula yields:
$$\sqrt {(\cos \alpha - \cos \beta )^2 + (\sin \alpha - \sin \beta )^2} = \sqrt {(\cos (\alpha - \beta ) - 1)^2 + (\sin (\alpha - \beta ) - 0)^2} $$
Squaring both sides to remove the square root:
$$(\cos \alpha - \cos \beta )^2 + (\sin \alpha - \sin \beta )^2 = (\cos (\alpha - \beta ) - 1)^2 + (\sin (\alpha - \beta ) - 0)^2$$
$$\cos ^2\alpha - 2\cos \alpha \cos \beta + \cos ^2\beta + \sin ^2\alpha - 2\sin \alpha \sin \beta + \sin ^2\beta = \cos ^2(\alpha - \beta ) - 2\cos (\alpha - \beta ) + 1 + \sin ^2(\alpha - \beta )$$
$$(\cos ^2\alpha + {\sin ^2}\alpha ) + (\cos ^2\beta + \sin ^2\beta ) - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = (\cos ^2(\alpha - \beta ) + \sin ^2(\alpha - \beta )) + 1 - 2\cos (\alpha - \beta )$$
$$2 - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = 2 - 2\cos (\alpha - \beta )$$
$$\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $$
So that's how it was derived. My questions are:
As the point P (the point where the terminal side of the angle $\alpha$ intersects the Unit Circle) is in the second quadrant, where $x<0$ and therefore $\cos (\alpha)<0$ shouldn't the coordinates for the point $P$ be $(-\cos(\alpha), \sin(\alpha))$? Furthermore what would happen if the angle $\alpha$was a quadrant III angle where both $\sin (\alpha)<0 $ and $\cos (\alpha )<)$. Surely this would affect computations of the cosine difference identity some what? If not, then why? Why does it seem like its computed only for quadrant I angles? Why does that identity stand for all quadrants is what I'm asking essentially?
The cosine difference formula is derived by equating the distance\length of the two chords $PQ$ and $AB$, however we "square out" the square root inherent in the distance formula. What if I were to do: $\sqrt {2 - 2\cos (\alpha - \beta )} $ (which is the simplified version of the RHS of the equation before we manipulate it to give us the identity) Would this give me the distance between the two chords?
What difference would it make if angle $\alpha $ was smaller than $\beta $?
What would happen if the angle $(\alpha -\beta)$ was greater $180^\circ$ or $\pi$ radians?
EDIT: Clarification of question 1 & 2., added a question 3 and 4.