Difference between revisions of "Arccos"
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[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] | ||
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Revision as of 05:52, 20 March 2015
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 $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:
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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: █
