Question Number 199447 by jabarsing last updated on 03/Nov/23
$${b}_{{n}} ={sin}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right)\Rightarrow\:{S}_{{n}} =? \\ $$
Answered by aleks041103 last updated on 03/Nov/23
$${b}_{{n}} ={Im}\left({e}^{{i}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right)} \right) \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{b}_{{k}} ={Im}\left({e}^{{ia}_{\mathrm{1}} } \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({e}^{{id}} \right)^{\left({k}−\mathrm{1}\right)} \right)= \\ $$$$={Im}\left({e}^{{ia}_{\mathrm{1}} } \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({e}^{{id}} \right)^{{k}} \right)={Im}\left({e}^{{ia}_{\mathrm{1}} } \frac{{e}^{{ind}} −\mathrm{1}}{{e}^{{id}} −\mathrm{1}}\right)= \\ $$$$={Im}\left({e}^{{ia}_{\mathrm{1}} } \frac{{e}^{{ind}/\mathrm{2}} \left({e}^{{ind}/\mathrm{2}} −{e}^{−{ind}/\mathrm{2}} \right)}{{e}^{{id}/\mathrm{2}} \left({e}^{{id}/\mathrm{2}} −{e}^{−{id}/\mathrm{2}} \right)}\right)= \\ $$$$={Im}\left({e}^{{i}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}/\mathrm{2}\right)} \frac{{sin}\left({nd}/\mathrm{2}\right)}{{sin}\left({d}/\mathrm{2}\right)}\right)= \\ $$$$=\frac{{sin}\left(\frac{{nd}}{\mathrm{2}}\right){sin}\left({a}_{\mathrm{1}} +\frac{\left({n}−\mathrm{1}\right){d}}{\mathrm{2}}\right)}{{sin}\left(\frac{{d}}{\mathrm{2}}\right)} \\ $$$$ \\ $$$$\Rightarrow{S}_{{n}} =\frac{{sin}\left(\frac{{nd}}{\mathrm{2}}\right){sin}\left({a}_{\mathrm{1}} +\frac{\left({n}−\mathrm{1}\right){d}}{\mathrm{2}}\right)}{{sin}\left(\frac{{d}}{\mathrm{2}}\right)} \\ $$
Commented by hardmath last updated on 03/Nov/23
$$\mathrm{perfect}\:\mathrm{solution}\:\mathrm{dear}\:\mathrm{professor} \\ $$