Question

$\sin x+\sin 3x+...+\sin (2n-1)x=\frac{{\sin }^{2}nx}{\sin x}$

Answer 1

Let P(n) be the given statement.
$$\begin{gathered} P\left(n\right):\sin x+\sin 3x+...+\sin \left(2n-1\right)x=\frac{{\sin }^{2}nx}{\sin x} \\ \mathrm{Step}1: \\ P\left(1\right):\sin x=\frac{{\sin }^{2}x}{\sin x} \\ \mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ \therefore \sin x+\sin 3x+...+\sin \left(2m-1\right)x=\frac{{\sin }^{2}mx}{\sin x} \\ \mathrm{We}\mathrm{shall}\mathrm{show}\mathrm{that}P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{We}\mathrm{know}\mathrm{that}P\left(m\right)\mathrm{is}\mathrm{true}. \\ \therefore \sin x+\sin 3x+...+\sin (2m-1)=\frac{{\sin }^{2}mx}{\sin x} \\ \Rightarrow \sin x+\sin 3x+...\sin (2m-1)x+\sin (2m+1)x=\frac{{\sin }^{2}mx}{\sin x}+\sin (2m+1)x\left(\mathrm{Adding}\sin (2m+1)x\mathrm{to}\mathrm{both}\mathrm{the}\mathrm{sides}\right) \\ \Rightarrow P(m+1)x=\frac{{\sin }^{2}mx+\sin x\left[\sin mx\cos \left(m+1\right)x+\sin \left(m+1\right)x\cos mx\right]}{\sin x} \\ =\frac{{\sin }^{2}mx+\sin x\left(\sin mx\cos mxcosx-{\sin }^{2}mx\sin x+\sin mx\cos x\cos mx+{\cos }^{2}mx\sin x\right)}{\sin x} \\ =\frac{{\sin }^{2}mx+2\sin x\cos x\cos mx-{\sin }^{2}x{\sin }^{2}mx+{\cos }^{2}mx{\sin }^{2}x}{\sin x} \\ =\frac{{\sin }^{2}mx\left(1-{\sin }^{2}x\right)+2\sin x\cos x\cos mx+{\cos }^{2}mx{\sin }^{2}x}{\sin x} \\ =\frac{{\sin }^{2}mx{\cos }^{2}x+2\sin x\cos x\cos mx+{\cos }^{2}mx{\sin }^{2}x}{\sin x} \\ =\frac{{\left(\sin mx\cos x+\cos mx\sin x\right)}^2}{\sin x} \\ =\frac{{\left[\sin \left(m+1\right)\right]}^2}{\sin x} \\ \mathrm{Hence},P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{By}\mathrm{the}\mathrm{principle}\mathrm{of}\mathrm{mathematical}\mathrm{induction},\mathrm{the}\mathrm{given}\mathrm{statement}P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$