Let \(P(\cos \alpha ; \sin \alpha)\) and \(Q(\cos \beta ; \sin \beta)\) be any two points on the unit circle \(O\) with radius 1.

If \(\mathrm{PO} \mathrm{X}=\alpha\) and \(\mathrm{QO} \mathrm{X}=\beta\) then \(\mathrm{PO} Q=\alpha-\beta\)

From the cosine rule:

\[

\begin{aligned}

& \mathrm{PQ}^2=1^2+1^2-2(1)(1) \cos (\alpha-\beta) \\

& \therefore \mathrm{PQ}^2=2-2 \cos (\alpha-\beta) \quad \ldots \mathrm{A}

\end{aligned}

\]

From the distance formula:

\[

\begin{aligned}

& \mathrm{PQ}^2=\left(x_{\mathrm{P}}-x_{\mathrm{Q}}\right)^2+\left(y_{\mathrm{P}}-y_{\mathrm{Q}}\right)^2 \\

& \therefore \mathrm{PQ}^2=(\cos \alpha-\cos \beta)^2+(\sin \alpha-\sin \beta)^2 \\

& \therefore \mathrm{PQ}^2=\cos ^2 \alpha-2 \cos \alpha \cdot \cos \beta+\cos ^2 \beta+\sin ^2 \alpha-2 \sin \alpha \cdot \sin \beta+\sin ^2 \beta \\

& \therefore \mathrm{PQ}^2=\cos ^2 \alpha+\sin ^2 \alpha+\cos ^2 \beta+\sin ^2 \beta-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta \\

& \therefore \mathrm{PQ}^2=1+1-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta \\

& \therefore \mathrm{PQ}^2=2-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta

\end{aligned}

\]

Now we can equate the two equations (A and B)

\[

\begin{array}{ll}

\therefore 2-2 \cos (\alpha-\beta)=2-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \quad(\mathrm{A}=\mathrm{B}) \\

\therefore-2 \cos (\alpha-\beta)=-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \\

\therefore \cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta & \div(-2)

\end{array}

\]