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 $y=\mathrm{arccos}(z)$ yields the formula <br />
+
Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula <br />
 
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
 
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
 
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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.

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 $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:

Sin(arccos(z)).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:

References

Weisstein, Eric W. "Inverse Cosine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseCosine.html