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Question Number 126701 by Eric002 last updated on 23/Dec/20

$$userighttrianglestoexplain$$$$why{cos}^{-1}\left(x\right)+{sin}^{-1}\left(x\right)=\pi /2$$

Answered by ebi last updated on 23/Dec/20

$$$$$$let{sin}^{-1}\left(x\right)=\theta ,thenx=sin\theta$$$$sin\theta =cos(\frac{\pi }{2}-\theta )$$$${cos}^{-1}\left(x\right)=\frac{\pi }{2}-\theta$$$${cos}^{-1}\left(x\right)=\frac{\pi }{2}-{sin}^{-1}\left(x\right)$$$${cos}^{-1}\left(x\right)+{sin}^{-1}\left(x\right)=\frac{\pi }{2}$$$$$$$$$$

Answered by mathmax by abdo last updated on 24/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arcosx}+\mathrm{arcsinx}-\frac{\pi }{2}$$$$\mathrm{\forall }\mathrm{x}\in [-1,1]{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-\frac{1}{\sqrt{1-{\mathrm{x}}^2}}+\frac{1}{\sqrt{1-{\mathrm{x}}^2}}=0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{c}$$$$\mathrm{f}\left(0\right)=\frac{\pi }{2}-\frac{\pi }{2}=0\Rightarrow \mathrm{\forall }\mathrm{x}\in [-1,1]\mathrm{f}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{arcosx}+\mathrm{arcsinx}=\frac{\pi }{2}$$

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