Prove that: sin²(n+1)A − sin²nA = sin(2n+1)A sin A

Question

Prove that:

\[ \sin^2(n+1)A-\sin^2 nA = \sin(2n+1)A\sin A \]

Proof

L.H.S.

\[ = \sin^2(n+1)A-\sin^2 nA \]

Using the identity:

\[ \sin^2 C-\sin^2 D = (\sin C-\sin D)(\sin C+\sin D) \]

\[ = [\sin(n+1)A-\sin nA] [\sin(n+1)A+\sin nA] \]

Using the identities:

\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \]

and

\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]

we get:

\[ = \left[ 2\cos\frac{(n+1)A+nA}{2} \sin\frac{(n+1)A-nA}{2} \right] \]

\[ \times \left[ 2\sin\frac{(n+1)A+nA}{2} \cos\frac{(n+1)A-nA}{2} \right] \]

\[ = \left[ 2\cos\frac{(2n+1)A}{2} \sin\frac{A}{2} \right] \]

\[ \times \left[ 2\sin\frac{(2n+1)A}{2} \cos\frac{A}{2} \right] \]

\[ = 4 \sin\frac{A}{2} \cos\frac{A}{2} \sin\frac{(2n+1)A}{2} \cos\frac{(2n+1)A}{2} \]

Using

\[ 2\sin x\cos x=\sin2x \]

\[ = \left(2\sin\frac{A}{2}\cos\frac{A}{2}\right) \left(2\sin\frac{(2n+1)A}{2}\cos\frac{(2n+1)A}{2}\right) \]

\[ = \sin A\sin(2n+1)A \]

\[ = \sin(2n+1)A\sin A \]

R.H.S.

\[ = \sin(2n+1)A\sin A \]

Hence,

\[ \sin^2(n+1)A-\sin^2 nA = \sin(2n+1)A\sin A \]

Hence proved.