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2-cos85-isin85-2-cos40-isin40-




Question Number 149121 by mathdanisur last updated on 03/Aug/21
((2∙(cos85° + isin85°))/( (√2)∙(cos40° + isin40°))) = ?
$$\frac{\mathrm{2}\centerdot\left({cos}\mathrm{85}°\:+\:\boldsymbol{{i}}{sin}\mathrm{85}°\right)}{\:\sqrt{\mathrm{2}}\centerdot\left({cos}\mathrm{40}°\:+\:\boldsymbol{{i}}{sin}\mathrm{40}°\right)}\:=\:? \\ $$
Answered by bramlexs22 last updated on 03/Aug/21
(√2) e^(i(85°−40°) )=(√2) e^(i45°) =(√2)(((√2)/2)+((√2)/2)i)     =1+i
$$\left.\sqrt{\mathrm{2}}\:\mathrm{e}^{\mathrm{i}\left(\mathrm{85}°−\mathrm{40}°\right.} \right)=\sqrt{\mathrm{2}}\:\mathrm{e}^{\mathrm{i45}°} =\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i}\right) \\ $$$$\:\:\:=\mathrm{1}+{i} \\ $$
Commented by mathdanisur last updated on 03/Aug/21
Ser, Thank You
$${Ser},\:{Thank}\:{You} \\ $$
Answered by Ar Brandon last updated on 03/Aug/21
=(√2)∙(e^(i85°) /e^(i40°) )=(√2)e^(i85°−i40°) =(√2)e^(i45°)   =(√2)(cos45°+isin45°)=1+i
$$=\sqrt{\mathrm{2}}\centerdot\frac{{e}^{{i}\mathrm{85}°} }{{e}^{{i}\mathrm{40}°} }=\sqrt{\mathrm{2}}{e}^{{i}\mathrm{85}°−{i}\mathrm{40}°} =\sqrt{\mathrm{2}}{e}^{{i}\mathrm{45}°} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{cos45}°+{i}\mathrm{sin45}°\right)=\mathrm{1}+{i} \\ $$
Commented by mathdanisur last updated on 03/Aug/21
Ser, Thank You
$${Ser},\:{Thank}\:{You} \\ $$

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