Difference between revisions of "Arccos"
(→Properties) |
|||
| Line 9: | Line 9: | ||
=Properties= | =Properties= | ||
| − | + | {{:Derivative of arccos}} | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
<div class="toccolours mw-collapsible mw-collapsed"> | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| Line 37: | Line 26: | ||
</div> | </div> | ||
</div> | </div> | ||
| + | |||
=References= | =References= | ||
[http://mathworld.wolfram.com/InverseCosine.html Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html] | [http://mathworld.wolfram.com/InverseCosine.html Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html] | ||
Revision as of 21:43, 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
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}},$$ where $\mathrm{arccos}$ denotes the inverse cosine function.
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
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
References
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: █
