Answer
If $A=B$, then $$\cos(A-B)=\cos A\cos B+\sin A\sin B=1$$
It agrees with the already known identity $\cos^2A+\sin^2A=1$
Work Step by Step
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
Since $A=B$, we replace $B$ with $A$, then we have:
- The left side: $\cos(A-B)=\cos(A-A)=\cos0=1$
- The right side: $$\cos A\cos B+\sin A\sin B=\cos A\cos A+\sin A\sin A=\cos^2A+\sin^2A=1$$
(Recall the identity $\cos^2A+\sin^2A=1$)
Therefore, if $A=B$, then both sides would equal $1$.
It agrees with the already known identity $\cos^2A+\sin^2A=1$
