Difference between revisions of "Arccos"
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The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br /> | 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 $ | + | Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula <br /> |
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$ | $$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$ | ||
</div> | </div> | ||
Revision as of 05:28, 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 $\theta=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
Proposition: $$\int \mathrm{arccos}(z) dz = z\mathrm{arccos}(z)-\sqrt{1-z^2}+C$$
Proof: █
Proposition: $$\mathrm{arccos}(z)=\mathrm{arcsec} \left( \dfrac{1}{z} \right)$$
Proof: █
