Question:
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
Solution:
LHS $=\sin ^{2}(n+1) A-\sin ^{2} n A$
$=\sin [(n+1) A+n A] \sin [(n+1) A-n A]$
$\left[\right.$ Using the formula $\sin ^{2} X-\sin ^{2} Y=\sin (X+Y) \sin (X-Y)$ and taking $X=(n+1) A$ and $\left.Y=n A\right]$
$=\sin [(n+1+n) A] \sin [(n+1-n) A]$
$=\sin (2 n+1) A \sin A$
= RHS
Hence proved.
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