Prove that:
(i) $\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=\frac{\sqrt{3}}{4}$
(ii) sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
(i) $\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}$
$=\cos \left(45^{\circ}+15^{\circ}\right) \cos \left(45^{\circ}-15^{\circ}\right) \quad\left[\cos ^{2} X-\sin ^{2} Y=\cos (X+Y) \cos (X-Y)\right]$
$=\cos 60^{\circ} \cos 30$
$=\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}$
Hence proved.
(ii) 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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