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Question Number 131879 by Study last updated on 09/Feb/21

$$provethatsin\left(n\pi \right)=0ifn\in \mathbb{Z}$$

Answered by physicstutes last updated on 10/Feb/21

$$\mathrm{prove}\mathrm{for}n=1$$$$\sin \pi =0$$$$\mathrm{assume}\mathrm{for}n=k,k\in \mathbb{Z}$$$$\Rightarrow \sin k\pi =0$$$$\mathrm{prove}\mathrm{for}n=k+1$$$$\Rightarrow \sin (k+1)\pi =\sin (k\pi +\pi )$$$$\sin (k\pi +\pi )=\sin k\pi \cos \pi +\cos k\pi \sin \pi$$$$\mathrm{but}\sin k\pi =0\mathrm{from}\mathrm{assumption}\mathrm{and}\sin \pi =0\mathrm{from}\mathrm{proved}\mathrm{step}$$$$\Rightarrow \sin (k+1)\pi =0$$$$\mathrm{hence}\mathrm{\forall }n\in \mathbb{Z}\sin \left(n\pi \right)=0$$

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