Question Number 196211 by pete last updated on 20/Aug/23
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}+\mathrm{i}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}+\frac{\mathrm{1}}{\mathrm{e}}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{e}−\frac{\mathrm{1}}{\mathrm{e}}\right)\mathrm{i} \\ $$
Answered by Frix last updated on 20/Aug/23
![cos x =((e^(2ix) +1)/(2e^(ix) )) cos (a+bi) =cos a cosh b −i sin a sinh b cos (π/3) cosh 1 −i sin (π/3) sinh 1 = =((cosh 1)/2)−(((√3)sinh 1)/2)= [cosh x =((e^(2x) +1)/(2e^x ))∧sinh x =((e^(2x) −1)/(2e^x ))] =((e^2 +1)/(4e))−(((√3)(e^2 −1))/(4e))i=(1/4)(e+(1/e))−((√3)/4)(e−(1/e))i](https://www.tinkutara.com/question/Q196212.png)
$$\mathrm{cos}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}}{\mathrm{2e}^{\mathrm{i}{x}} } \\ $$$$\mathrm{cos}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{cos}\:{a}\:\mathrm{cosh}\:{b}\:−\mathrm{i}\:\mathrm{sin}\:{a}\:\mathrm{sinh}\:{b} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\:\mathrm{cosh}\:\mathrm{1}\:−\mathrm{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\:\mathrm{sinh}\:\mathrm{1}\:= \\ $$$$=\frac{\mathrm{cosh}\:\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}\mathrm{sinh}\:\mathrm{1}}{\mathrm{2}}= \\ $$$$\left[\mathrm{cosh}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{2e}^{{x}} }\wedge\mathrm{sinh}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{2e}^{{x}} }\right] \\ $$$$=\frac{\mathrm{e}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4e}}−\frac{\sqrt{\mathrm{3}}\left(\mathrm{e}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4e}}\mathrm{i}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}+\frac{\mathrm{1}}{\mathrm{e}}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{e}−\frac{\mathrm{1}}{\mathrm{e}}\right)\mathrm{i} \\ $$
Commented by pete last updated on 20/Aug/23
$$\mathrm{You}\:\mathrm{do}\:\mathrm{all}\:\mathrm{sir},\:\mathrm{many}\:\mathrm{thanks}\:\mathrm{sir}. \\ $$
Commented by Frix last updated on 20/Aug/23
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