求证:(cosπ/6+isinπ/6)^n=cos(nπ/6)+i *sin( nπ/6)谢谢
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求证:(cosπ/6+isinπ/6)^n=cos(nπ/6)+i *sin( nπ/6)谢谢
证明:
当n=1时,(cosπ/6+isinπ/6)^1=cos(π/6)+i sin( π/6)显然成立
假设n=k,(cosπ/6+isinπ/6)^k=cos(kπ/6)+i *sin( kπ/6)成立,
当n=k+1时,(cosπ/6+isinπ/6)^(k+1)
=(cosπ/6+isinπ/6)^k*(cosπ/6+isinπ/6)
=(coskπ/6+isin kπ/6)*(cosπ/6+isinπ/6)
=coskπ/6*cosπ/6-sinkπ/6sinπ/6+i(sinπ/6coskπ/6+sinkπ/6cosπ/6)
=cos[(k+1)π/6]+isin[(k+1)π/6]
证毕.
当n=1时,(cosπ/6+isinπ/6)^1=cos(π/6)+i sin( π/6)显然成立
假设n=k,(cosπ/6+isinπ/6)^k=cos(kπ/6)+i *sin( kπ/6)成立,
当n=k+1时,(cosπ/6+isinπ/6)^(k+1)
=(cosπ/6+isinπ/6)^k*(cosπ/6+isinπ/6)
=(coskπ/6+isin kπ/6)*(cosπ/6+isinπ/6)
=coskπ/6*cosπ/6-sinkπ/6sinπ/6+i(sinπ/6coskπ/6+sinkπ/6cosπ/6)
=cos[(k+1)π/6]+isin[(k+1)π/6]
证毕.
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