Question

Prove that:
(i) ${\cos }^{2}45°-{\sin }^{2}15°=\frac{\sqrt{3}}{4}$
(ii) sin² (n + 1) A − sin² nA = sin (2n + 1) A sin A.

Answer 1

(i)
$$\begin{gathered} {\cos }^{2}45°-{\sin }^{2}15° \\ =\cos \left(45°+15°\right)\cos \left(45°-15°\right)\left[{\cos }^{2}X-{\sin }^{2}Y=\cos \left(X+Y\right)\cos \left(X-Y\right)\right] \\ =\cos 60°\cos 30° \\ =\frac{1}{2}\times \frac{\sqrt{3}}{2} \\ =\frac{\sqrt{3}}{4} \end{gathered}$$

Hence proved.

(ii)
$$\begin{gathered} \mathrm{LHS}={\sin }^2\left(n+1\right)A-{\sin }^{2}nA \\ =\sin \left[\left(n+1\right)A+nA\right]\sin \left[\left(n+1\right)A-nA\right]\left[\text{U}\mathrm{sing}\mathrm{the}\mathrm{formula}{\sin }^{2}X-{\sin }^{2}Y=\sin \left(X+Y\right)\sin \left(X-Y\right) \\ \mathrm{and}\mathrm{taking}X=\left(n+1\right)A\mathrm{and}Y=nA\right] \\ =\sin \left[\left(n+1+n\right)A\right]\sin \left[\left(n+1-n\right)A\right] \\ =\sin \left(2n+1\right)A\sin A \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$