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Question Number 195157 by Erico last updated on 25/Jul/23

$$\mathrm{Prove}\mathrm{that}$$$$\frac{x^3}{2{sin}^2\left(\frac{1}{2}arctan\frac{x}{y}\right)}+\frac{y^3}{2{cos}^2\left(\frac{1}{2}arctan\frac{y}{x}\right)}=(x+y)(x^2+y^2)$$

Answered by Frix last updated on 25/Jul/23

$${\sin }^2{\tan }^{-1}\alpha =\frac{\sqrt{{\alpha }^2+1}-1}{2\sqrt{{\alpha }^2+1}}=\frac{\sqrt{x^2+y^2}-y}{2\sqrt{x^2+y^2}}$$$${\cos }^2{\tan }^{-1}\frac{1}{\alpha }=\frac{\sqrt{{\alpha }^2+1}+\alpha }{2\sqrt{{\alpha }^2+1}}=\frac{\sqrt{x^2+y^2}+x}{2\sqrt{x^2+y^2}}$$$$\frac{x^3}{2\frac{\sqrt{x^2+y^2}-y}{2\sqrt{x^2+y^2}}}+\frac{y^3}{2\frac{\sqrt{x^2+y^2}+x}{2\sqrt{x^2+y^2}}}=$$$$=\sqrt{x^2+y^2}(\frac{x^3}{\sqrt{x^2+y^2}-y}+\frac{y^3}{\sqrt{x^2+y^2}+x})=$$$$=\sqrt{x^2+y^2}(x(\sqrt{x^2+y^2}+y)+y(\sqrt{x^2+y^2}-x))=$$$$=\sqrt{x^2+y^2}(x+y)\sqrt{x^2+y^2}=$$$$=(x+y)(x^2+y^2)$$

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