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

$$soitz\in \mathbb{C}montrerquecos\left(z\right)etsin\left(z\right)$$$$nesontpasbornees$$$$quevaut{sin}^2\left(z\right)+{cos}^2\left(z\right)=??$$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

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

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