Difference between revisions of "Arccos"
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$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$$ | $$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$$ | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> If $ | + | <strong>Proof:</strong> If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get |
| − | $$-\sin( | + | $$-\sin(\theta)\theta'=1.$$ |
| − | + | The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br /> | |
[[File:Sin(arccos(z)).png|center|200px]] | [[File:Sin(arccos(z)).png|center|200px]] | ||
Hence substituting back in $y=\mathrm{arccos}(z)$ yields the formula <br /> | Hence substituting back in $y=\mathrm{arccos}(z)$ yields the formula <br /> | ||
Revision as of 05:21, 31 October 2014
The function $\mathrm{arccos} \colon [-1,1] \longrightarrow [0,\pi]$ is the inverse function of the cosine function.
Graph of $\mathrm{arccos}$ on $[-1,1]$.
Properties
Proposition: $$\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.$$
The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:
).png)
Hence substituting back in $y=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
Proposition: $$\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: █
