Question:
If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.
Solution:
$\mathrm{RHS}=m^{2}+n^{2}-2$
$=(\cos A+\sin B)^{2}+(\sin A+\cos B)^{2}-2$
$=\cos ^{2} A+\sin ^{2} B+2 \cos A \sin B+\sin ^{2} A+\cos ^{2} B+2 \sin A \cos B-2$
$=1+1+2 \cos A \sin B+2 \sin A \cos B-2$
$=2(\cos A \sin B+\sin A \cos B)$
$=2 \sin (A+B)$
= RHS
Hence proved.