Prove that: sin²B = sin²A + sin²(A−B) − 2 sinA cosB sin(A−B)

Question

Prove that:

\[ \sin^2 B = \sin^2 A + \sin^2(A-B) – 2\sin A\cos B\sin(A-B) \]

Proof

R.H.S.

\[ = \sin^2 A+\sin^2(A-B)-2\sin A\cos B\sin(A-B) \]

\[ = \sin^2 A+(\sin A\cos B-\cos A\sin B)^2 \]

\[ -2\sin A\cos B(\sin A\cos B-\cos A\sin B) \]

\[ = \sin^2 A+\sin^2 A\cos^2 B+\cos^2 A\sin^2 B \]

\[ -2\sin A\cos A\sin B\cos B \]

\[ -2\sin^2 A\cos^2 B +2\sin A\cos A\sin B\cos B \]

\[ = \sin^2 A-\sin^2 A\cos^2 B+\cos^2 A\sin^2 B \]

\[ = \sin^2 A(1-\cos^2 B)+\cos^2 A\sin^2 B \]

\[ = \sin^2 A\sin^2 B+\cos^2 A\sin^2 B \]

\[ = \sin^2 B(\sin^2 A+\cos^2 A) \]

\[ = \sin^2 B \]

L.H.S. = R.H.S.

Hence proved.