Prove that sin

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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