Prove that:
Question:

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.

Solution:

(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.