Difference between revisions of "Arccos"

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The function $\mathrm{arccos} \colon [-1,1] \longrightarrow [0,\pi]$ is the inverse function of the [[cosine]] function.
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__NOTOC__
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The function $\mathrm{arccos} \colon \mathbb{C} \setminus \{(-\infty,-1) \bigcup (1,\infty) \} \rightarrow \mathbb{C}$ is defined by
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$$\rm{arccos}(z)=\dfrac{\pi}{2} + i\log\left( iz + \sqrt{1-z^2} \right),$$
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where $i$ denotes the [[imaginary number]] and $\log$ denotes the [[logarithm]].
  
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[[File:Arccos.png|500px]]
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<div align="center">
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<gallery>
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File:Arccosplot.png|Graph of $\mathrm{arccos}$ on $[-1,1]$.
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File:Complexarccosplot.png|[[Domain coloring]] of $\mathrm{arccos}$.
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</gallery>
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</div>
  
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[[File:Complex arccos.jpg|500px]]
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=Properties=
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[[Arccos as inverse cosine]]<br />
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[[Derivative of arccos]]<br />
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[[Antiderivative of arccos]]<br />
  
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=Properties=
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=References=
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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[http://mathworld.wolfram.com/InverseCosine.html  Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html]
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<strong>Proposition:</strong>
 
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$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$$
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> If $y=\mathrm{arccos}(z)$ then $\cos(y)=z$. Now use [[implicit differentiation]] with respect to $z$ to get
 
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$$-\sin(y)y'=1.$$
 
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Substituting back in $y=\mathrm{arccos}(z)$ yields the formula
 
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$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$ 
 
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</div>
 
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</div>
 
  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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=See Also=
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<strong>Proposition:</strong>
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[[Cosine]] <br />
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$$\int \mathrm{arccos}(z) dz = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C$$
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[[Cosh]] <br />
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<div class="mw-collapsible-content">
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[[Arccosh]]
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
  
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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{{:Inverse trigonometric functions footer}}
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<strong>Proposition:</strong>
 
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$$\mathrm{arccos}(z)=\mathrm{arcsec} \left( \dfrac{1}{z} \right)$$
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
  
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=References=
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[[Category:SpecialFunction]]
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[http://mathworld.wolfram.com/InverseCosine.html  Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html]
 

Latest revision as of 20:04, 22 November 2016

The function $\mathrm{arccos} \colon \mathbb{C} \setminus \{(-\infty,-1) \bigcup (1,\infty) \} \rightarrow \mathbb{C}$ is defined by $$\rm{arccos}(z)=\dfrac{\pi}{2} + i\log\left( iz + \sqrt{1-z^2} \right),$$ where $i$ denotes the imaginary number and $\log$ denotes the logarithm.

  • Graph of $\mathrm{arccos}$ on $[-1,1]$.

  • Domain coloring of $\mathrm{arccos}$.

Properties

Arccos as inverse cosine
Derivative of arccos
Antiderivative of arccos

References

Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html

See Also

Cosine
Cosh
Arccosh

Inverse trigonometric functions

Arcsin

Arccos

Arctan

Arccsc

Arcsec

Arccot
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