Difference between revisions of "Arccos"
| Line 4: | Line 4: | ||
<gallery> | <gallery> | ||
File:Arccos.png|Graph of $\mathrm{arccos}$ on $[-1,1]$. | File:Arccos.png|Graph of $\mathrm{arccos}$ on $[-1,1]$. | ||
| − | File: | + | File:Complexarccosplot.png|[[Domain coloring]] of $\mathrm{arccos}$. |
</gallery> | </gallery> | ||
</div> | </div> | ||
Revision as of 21:42, 15 May 2016
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 $\mathrm{arccos}$.
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: █
