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Question Number 145442 by ArielVyny last updated on 04/Jul/21

soit z∈C montrer que cos(z) et sin (z) ne sont pas bornees que vaut sin^2 (z)+cos^2 (z)=??

$${soit}\:{z}\in\mathbb{C}\:{montrer}\:{que}\:{cos}\left({z}\right)\:{et}\:{sin}\:\left({z}\right) \\ $$$${ne}\:{sont}\:{pas}\:{bornees} \\ $$$${que}\:{vaut}\:{sin}^{\mathrm{2}} \left({z}\right)+{cos}^{\mathrm{2}} \left({z}\right)=?? \\ $$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

cos(iy) = ((e^(i(iy)) +e^(−i(iy)) )/2) = ((e^(−y) +e^y )/2) = chy sin(iy) = ((e^(i(iy)) −e^(−i(iy)) )/(2i)) = ((e^(−y) −e^y )/(2i)) = ishy cosz = cos(x+iy) cosz = cosxcos(iy)−sinxsin(iy) cosz = cosxchy−isinxshy cosz non borne a cause de chy et shy sinz = sin(x+iy) sinz = sinxcos(iy)+cosxsin(iy) sinz = sinxchy+icosxshy sinz non borne a cause de chy et shy cos^2 z = (cosxchy−isinxshy)^2 cos^2 z = cos^2 xch^2 y−sin^2 xsh^2 y −2icosxsinxchyshy sin^2 z = (sinxchy+icosxshy)^2 sin^2 z = sin^2 xch^2 y−cos^2 xsh^2 y +2icosxsinxchyshy cos^2 z+sin^2 z = cos^2 xch^2 y−sin^2 xsh^2 y +sin^2 xch^2 y−cos^2 xsh^2 y cos^2 z+sin^2 z = (ch^2 y−sh^2 y)(cos^2 x+sin^2 x) cos^2 z+sin^2 z = 1×1 = 1 On retrouve la meme relation fondamentale que pour les reels.

$$\mathrm{cos}\left({iy}\right)\:=\:\frac{{e}^{{i}\left({iy}\right)} +{e}^{−{i}\left({iy}\right)} }{\mathrm{2}}\:=\:\frac{{e}^{−{y}} +{e}^{{y}} }{\mathrm{2}}\:=\:\mathrm{ch}{y} \\ $$$$\mathrm{sin}\left({iy}\right)\:=\:\frac{{e}^{{i}\left({iy}\right)} −{e}^{−{i}\left({iy}\right)} }{\mathrm{2}{i}}\:=\:\frac{{e}^{−{y}} −{e}^{{y}} }{\mathrm{2}{i}}\:=\:{i}\mathrm{sh}{y} \\ $$$$ \\ $$$$\mathrm{cos}{z}\:=\:\mathrm{cos}\left({x}+{iy}\right) \\ $$$$\mathrm{cos}{z}\:=\:\mathrm{cos}{x}\mathrm{cos}\left({iy}\right)−\mathrm{sin}{x}\mathrm{sin}\left({iy}\right) \\ $$$$\mathrm{cos}{z}\:=\:\mathrm{cos}{x}\mathrm{ch}{y}−{i}\mathrm{sin}{x}\mathrm{sh}{y} \\ $$$$\mathrm{cos}{z}\:\mathrm{non}\:\mathrm{borne}\:\mathrm{a}\:\mathrm{cause}\:\mathrm{de}\:\mathrm{ch}{y}\:\mathrm{et}\:\mathrm{sh}{y} \\ $$$$ \\ $$$$\mathrm{sin}{z}\:=\:\mathrm{sin}\left({x}+{iy}\right) \\ $$$$\mathrm{sin}{z}\:=\:\mathrm{sin}{x}\mathrm{cos}\left({iy}\right)+\mathrm{cos}{x}\mathrm{sin}\left({iy}\right) \\ $$$$\mathrm{sin}{z}\:=\:\mathrm{sin}{x}\mathrm{ch}{y}+{i}\mathrm{cos}{x}\mathrm{sh}{y} \\ $$$$\mathrm{sin}{z}\:\mathrm{non}\:\mathrm{borne}\:\mathrm{a}\:\mathrm{cause}\:\mathrm{de}\:\mathrm{ch}{y}\:\mathrm{et}\:\mathrm{sh}{y} \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} {z}\:=\:\left(\mathrm{cos}{x}\mathrm{ch}{y}−{i}\mathrm{sin}{x}\mathrm{sh}{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{2}} {z}\:=\:\mathrm{cos}^{\mathrm{2}} {x}\mathrm{ch}^{\mathrm{2}} {y}−\mathrm{sin}^{\mathrm{2}} {x}\mathrm{sh}^{\mathrm{2}} {y} \\ $$$$−\mathrm{2}{i}\mathrm{cos}{x}\mathrm{sin}{x}\mathrm{ch}{y}\mathrm{sh}{y} \\ $$$$ \\ $$$$\mathrm{sin}^{\mathrm{2}} {z}\:=\:\left(\mathrm{sin}{x}\mathrm{ch}{y}+{i}\mathrm{cos}{x}\mathrm{sh}{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} {z}\:=\:\mathrm{sin}^{\mathrm{2}} {x}\mathrm{ch}^{\mathrm{2}} {y}−\mathrm{cos}^{\mathrm{2}} {x}\mathrm{sh}^{\mathrm{2}} {y} \\ $$$$+\mathrm{2}{i}\mathrm{cos}{x}\mathrm{sin}{x}\mathrm{ch}{y}\mathrm{sh}{y} \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} {z}+\mathrm{sin}^{\mathrm{2}} {z}\:=\:\mathrm{cos}^{\mathrm{2}} {x}\mathrm{ch}^{\mathrm{2}} {y}−\mathrm{sin}^{\mathrm{2}} {x}\mathrm{sh}^{\mathrm{2}} {y} \\ $$$$+\mathrm{sin}^{\mathrm{2}} {x}\mathrm{ch}^{\mathrm{2}} {y}−\mathrm{cos}^{\mathrm{2}} {x}\mathrm{sh}^{\mathrm{2}} {y} \\ $$$$\mathrm{cos}^{\mathrm{2}} {z}+\mathrm{sin}^{\mathrm{2}} {z}\:=\:\left(\mathrm{ch}^{\mathrm{2}} {y}−\mathrm{sh}^{\mathrm{2}} {y}\right)\left(\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} {z}+\mathrm{sin}^{\mathrm{2}} {z}\:=\:\mathrm{1}×\mathrm{1}\:=\:\mathrm{1} \\ $$$$\mathrm{On}\:\mathrm{retrouve}\:\mathrm{la}\:\mathrm{meme}\:\mathrm{relation} \\ $$$$\mathrm{fondamentale}\:\mathrm{que}\:\mathrm{pour}\:\mathrm{les}\:\mathrm{reels}. \\ $$

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