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Question Number 48225 by gunawan last updated on 21/Nov/18
prove that  exp(((2+πi)/4))=(√(e/2))(1+i)  cos (z_1 +z_2 )=cos z_1 cos z_2 −sin z_1 sin z_2
$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{exp}\left(\frac{\mathrm{2}+\pi\mathrm{i}}{\mathrm{4}}\right)=\sqrt{\frac{{e}}{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{cos}\:\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)=\mathrm{cos}\:{z}_{\mathrm{1}} \mathrm{cos}\:{z}_{\mathrm{2}} −\mathrm{sin}\:{z}_{\mathrm{1}} \mathrm{sin}\:{z}_{\mathrm{2}} \\ $$
Answered by Smail last updated on 21/Nov/18
e^((2+πi)/4) =e^((1/2)+i(π/4)) =(√e)×e^(iπ/4) =(√e)×(cos((π/4))+isin((π/4)))  =(√e)(((√2)/2)+i((√2)/2))=((√e)/( (√2)))(1+i)  =(√(e/2))(1+i)
$${e}^{\frac{\mathrm{2}+\pi{i}}{\mathrm{4}}} ={e}^{\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\pi}{\mathrm{4}}} =\sqrt{{e}}×{e}^{{i}\pi/\mathrm{4}} =\sqrt{{e}}×\left({cos}\left(\frac{\pi}{\mathrm{4}}\right)+{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\sqrt{{e}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\sqrt{{e}}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$$=\sqrt{\frac{{e}}{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$

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