Résoudre tan^-1(1)=x | Microsoft Math Solver

Calculer x

x = \frac{200000000000000000}{3491012985643517} = 57\frac{1012259818319531}{3491012985643517} \approx 57,289961631

Étapes de calcul des équations linéaires

\tan ^ { - 1 } ( 1 ) = x

Évaluer les fonctions trigonométriques du problème

0,017455064928217585 ^ {-1} = x

Calculer 0,017455064928217585 à la puissance -1 et obtenir \frac{200000000000000000}{3491012985643517}.

\frac{200000000000000000}{3491012985643517}=x

Échanger les côtés afin que tous les termes de variable soient à gauche.

x=\frac{200000000000000000}{3491012985643517}

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Trigonometry

5 problèmes semblables à :

\tan ^ { - 1 } ( 1 ) = x

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Transform the function f(x) =\tan^2 x into Fourier series for -\frac{\pi}{2}\leq x \leq \frac{\pi}{2}

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Dear, Anon... God made so many brilliant methods, why you want it via first principle? \displaystyle f(x)=x\tan ^{-1}x \displaystyle f'(1)=\lim_{h\rightarrow 0}\left [ \frac{f(1+h)-f(1)}{h} \right ] ...

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\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}} ...

How to evaluate \int 1/(1+x^{2n})\,dx for an arbitrary positive integer n?

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If the integral is taken from 0 to \infty, there is more than one way to evaluate this. One is \begin{align} \int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}} &=\int_0^1\frac{\mathrm{d}t}{1+t^{2n}}+\int_0^1\frac{t^{2n-2}\,\mathrm{d}t}{1+t^{2n}}\\ &=\int_0^1(1-t^{2n}+t^{4n}-t^{6n}+\dots)\,\mathrm{d}t\\ &+\int_0^1(t^{2n-2}-t^{4n-2}+t^{6n-2}+\dots)\,\mathrm{d}t\\ &=1-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{6n+1}+\dots\\ &+\frac{1}{2n-1}-\frac{1}{4n-1}+\frac{1}{6n-1}-\dots\\ &=\frac{1}{2n}\left(\frac{1}{0+\frac{1}{2n}}-\frac{1}{1+\frac{1}{2n}}+\frac{1}{2+\frac{1}{2n}}-\frac{1}{3+\frac{1}{2n}}+\dots\right)\\ &+\frac{1}{2n}\left(-\frac{1}{-1+\frac{1}{2n}}+\frac{1}{-2+\frac{1}{2n}}-\frac{1}{-3+\frac{1}{2n}}-\dots\right)\\ &=\frac{1}{2n}\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac{1}{2n}}\\ &=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{1} \end{align} ...

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0,017455064928217585 ^ {-1} = x

Évaluer les fonctions trigonométriques du problème

\frac{200000000000000000}{3491012985643517}=x

Calculer 0,017455064928217585 à la puissance -1 et obtenir \frac{200000000000000000}{3491012985643517}.

x=\frac{200000000000000000}{3491012985643517}

Échanger les côtés afin que tous les termes de variable soient à gauche.

Exemples

Équation du second degré

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonométrie

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Équation simultanée

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Différenciation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Intégration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Limites

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

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