Difference between revisions of "Arccos"
| Line 11: | Line 11: | ||
<div class="toccolours mw-collapsible mw-collapsed"> | <div class="toccolours mw-collapsible mw-collapsed"> | ||
<strong>Proposition:</strong> | <strong>Proposition:</strong> | ||
| − | + | $\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$ | |
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | <strong>Proof:</strong> If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | ||
| Line 24: | Line 24: | ||
<div class="toccolours mw-collapsible mw-collapsed"> | <div class="toccolours mw-collapsible mw-collapsed"> | ||
<strong>Proposition:</strong> | <strong>Proposition:</strong> | ||
| − | + | $\displaystyle\int \mathrm{arccos}(z) dz = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C$ | |
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
| Line 32: | Line 32: | ||
<div class="toccolours mw-collapsible mw-collapsed"> | <div class="toccolours mw-collapsible mw-collapsed"> | ||
<strong>Proposition:</strong> | <strong>Proposition:</strong> | ||
| − | + | $\mathrm{arccos}(z)=\mathrm{arcsec} \left( \dfrac{1}{z} \right)$ | |
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Revision as of 05:58, 31 October 2014
The function $\mathrm{arccos} \colon [-1,1] \longrightarrow [0,\pi]$ is the inverse function of the cosine function.
Graph of $\mathrm{arccos}$ on $[-1,1]$.
Domain coloring of analytic continuation of $\mathrm{arccos}$.
Properties
Proposition: $\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$
Proof: If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use implicit differentiation with respect to $z$ to get
$$-\sin(\theta)\theta'=1.$$
The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:
).png)
Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
Proposition: $\displaystyle\int \mathrm{arccos}(z) dz = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C$
Proof: █
Proposition: $\mathrm{arccos}(z)=\mathrm{arcsec} \left( \dfrac{1}{z} \right)$
Proof: █
