Question Number 35223 by abdo mathsup 649 cc last updated on 16/May/18
$${what}\:{is}\:{the}\:{value}\:{of}\:{cos}\left({i}+{j}\right)\:{with}\:{i}^{\mathrm{2}} =−\mathrm{1}\:{and} \\ $$$${j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:? \\ $$
Answered by sma3l2996 last updated on 17/May/18
$${cos}\left({i}+{j}\right)=\frac{{e}^{\left({i}+{j}\right){i}} +{e}^{−\left({i}+{j}\right){i}} }{\mathrm{2}}=\frac{{e}^{−\mathrm{1}+{ij}} +{e}^{\mathrm{1}−{ij}} }{\mathrm{2}} \\ $$$${ij}={i}×{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} ={i}\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left({i}+{j}\right)=\frac{{e}^{−\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}} +{e}^{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{i}\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}=\frac{{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}} ×{e}^{{i}/\mathrm{2}} +\frac{{e}^{−{i}/\mathrm{2}} }{{e}^{\sqrt{\mathrm{3}}/\mathrm{2}−\mathrm{1}} }}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{e}^{\sqrt{\mathrm{3}}/\mathrm{2}−\mathrm{1}} }\left({e}^{\sqrt{\mathrm{3}}−\mathrm{2}} \left({cos}\frac{\mathrm{1}}{\mathrm{2}}+{isin}\frac{\mathrm{1}}{\mathrm{2}}\right)+{cos}\frac{\mathrm{1}}{\mathrm{2}}−{isin}\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{e}^{\sqrt{\mathrm{3}}/\mathrm{2}−\mathrm{1}} }\left(\left({e}^{\sqrt{\mathrm{3}}−\mathrm{2}} +\mathrm{1}\right){cos}\left(\mathrm{1}/\mathrm{2}\right)+{i}\left({e}^{\sqrt{\mathrm{3}}−\mathrm{2}} −\mathrm{1}\right){sin}\left(\mathrm{1}/\mathrm{2}\right)\right) \\ $$
Commented by abdo imad last updated on 17/May/18
$${sir}\:{Sma}\mathrm{3}{l}\:{your}\:{method}\:{is}\:{correct}\:{but}\:{you}\:{have}\:{commited} \\ $$$${a}\:{small}\:{error}\:\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:!… \\ $$