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Question-138594

Question Number 138594 by Ar Brandon last updated on 15/Apr/21 Commented by Ar Brandon last updated on 15/Apr/21 $$\mathrm{In}\:\mathrm{honour}\:\mathrm{of}\:\mathrm{Leonhard}\:\mathrm{Euler} \\ $$$$\mathrm{on}\:\mathrm{his}\:\mathrm{314}^{\mathrm{th}} \:\mathrm{anniversary}. \\ $$ Commented…

A-2-3-rectangle-and-a-3-4-rectangle-are-contain-within-a-square-without-over-laping-at-any-inferior-point-and-the-sides-of-the-square-are-parallel-to-the-sides-of-the-two-given-rectangles-

Question Number 7519 by Tawakalitu. last updated on 01/Sep/16 $${A}\:\left(\mathrm{2}\:×\:\mathrm{3}\right)\:{rectangle}\:{and}\:{a}\:\left(\mathrm{3}\:×\:\mathrm{4}\right)\:{rectangle}\:{are}\:{contain}\:{within}\: \\ $$$${a}\:{square}\:{without}\:{over}\:{laping}\:{at}\:{any}\:{inferior}\:{point}\:,\:{and}\:{the}\: \\ $$$${sides}\:{of}\:{the}\:{square}\:{are}\:{parallel}\:{to}\:{the}\:{sides}\:{of}\:{the}\:{two}\:{given} \\ $$$${rectangles}.\:{what}\:{is}\:{the}\:{smallest}\:{possible}\:{area}\:{of}\:{the}\:{square}. \\ $$ Commented by Rasheed Soomro last updated on…

let-P-n-X-n-X-n-1-X-2-X-1-R-X-1-prove-that-P-n-have-one-root-x-n-inside-0-2-study-the-sequence-x-n-

Question Number 73052 by mathmax by abdo last updated on 05/Nov/19 $${let}\:{P}_{{n}} ={X}^{{n}} \:+{X}^{{n}−\mathrm{1}} \:+….+{X}^{\mathrm{2}} \:+{X}−\mathrm{1}\:\in{R}\left[{X}\right] \\ $$$$\left.\mathrm{1}\left.\right){prove}\:{that}\:{P}_{{n}} {have}\:{one}\:{root}\:{x}_{{n}} \:{inside}\:\right]\mathrm{0},+\infty\left[\right. \\ $$$$\left.\mathrm{2}\right){study}\:{the}\:{sequence}\:{x}_{{n}} \\ $$ Answered…

Question-7514

Question Number 7514 by Tawakalitu. last updated on 01/Sep/16 Answered by Yozzia last updated on 01/Sep/16 $${sin}\left(\pi{cos}\alpha\right)={cos}\left(\pi{sin}\alpha\right). \\ $$$${Using}\:{sina}={cos}\left(\frac{\pi}{\mathrm{2}}−{a}\right),\:{we}\:{get} \\ $$$${cos}\left(\frac{\pi}{\mathrm{2}}−\pi{cos}\alpha\right)={cos}\left(\pi{sin}\alpha\right) \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}−\pi{cos}\alpha=\mathrm{2}{n}\pi\pm\pi{sin}\alpha\:\:\:\:\left({n}\in\mathbb{Z}\right) \\ $$$$\mp\pi{sin}\alpha−\pi{cos}\alpha=\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{2}}…