Question

Prove that sin² (n + 1) A − sin² nA = sin (2n + 1) A sin A.

Answer 1

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