Question

Prove that the identity \(\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \cdot \cos \mathrm{B}+\sin \mathrm{A} \cdot \sin \mathrm{B}\)

Answer 1

 

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)
\[
\begin{array}{ll}
\therefore 2-2 \cos (\alpha-\beta)=2-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \quad(\mathrm{A}=\mathrm{B}) \\
\therefore-2 \cos (\alpha-\beta)=-2 \cos \alpha \cdot \cos \beta-2 \sin \alpha \cdot \sin \beta & \\
\therefore \cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta & \div(-2)
\end{array}
\]