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Prove that the identity $\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \cdot \cos \mathrm{B}+\sin \mathrm{A} \cdot \sin \mathrm{B}$
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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)
$\begin{array}{ll} \therefore 2-2 \cos (\alpha-\beta)=2-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \quad(\mathrm{A}=\mathrm{B}) \\ \therefore-2 \cos (\alpha-\beta)=-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \\ \therefore \cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta & \div(-2) \end{array}$

by Diamond (75,025 points)

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