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

Datrys ar gyfer x

x = \frac{200000000000000000}{3491012985643517} = 57\frac{1012259818319531}{3491012985643517} \approx 57.289961631

Camau ar gyfer Datrys Hafaliad Llinol

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

Evaluate trigonometric functions in the problem

0.017455064928217585 ^ {-1} = x

Cyfrifo 0.017455064928217585 i bŵer -1 a chael \frac{200000000000000000}{3491012985643517}.

\frac{200000000000000000}{3491012985643517}=x

Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.

x=\frac{200000000000000000}{3491012985643517}

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Cwis

Trigonometry

5 problemau tebyg i:

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

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Truong-Son N. Jul 5, 2015 Assuming you don't remember the derivative of \displaystyle{\arctan{{u}}} : \displaystyle{y}={{\arctan{{x}}}^{{2}}} \displaystyle{\tan{{y}}}={x}^{{2}} ...

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\displaystyle\frac{{1}}{{x}}=\frac{{1}}{{{\tan}^{{-{1}}}{t}}} Explanation: \displaystyle\frac{{1}}{{x}}=\frac{{1}}{{{\tan}^{{-{1}}}{t}}} Note that \displaystyle\frac{{1}}{{x}}={x}^{{-{1}}} ...

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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}} ...

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https://math.stackexchange.com/questions/171024/how-to-evaluate-int-1-1x2n-dx-for-an-arbitrary-positive-integer-n

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

Evaluate trigonometric functions in the problem

\frac{200000000000000000}{3491012985643517}=x

Cyfrifo 0.017455064928217585 i bŵer -1 a chael \frac{200000000000000000}{3491012985643517}.

x=\frac{200000000000000000}{3491012985643517}

Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.

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