Question

${(\cos A-\cos B)}^2+{(\sin A-\sin B)}^2=c{\sin }^2(A-B)/2$
Find $2c$

Answer 1

L.H.S. $={(\cos A-\cos B)}^2+{(\sin A-\sin B)}^2$
$={\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B+{\sin }^{2}A+{\sin }^{2}B-2\sin A\sin B$

$=({\cos }^{2}A+{\sin }^{2}A)+({\cos }^{2}B+{\sin }^{2}B)-2(\cos A\cos B+\sin A\sin B)$

$=2-2\cos (A-B)=2(1-\cos (A-B))=4{\sin }^2(A-B)/2$

Therefore, $2c=8$
Ans: 8