Difference between revisions of "Arccos"

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(Properties)
(Properties)
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$$-\sin(y)y'=1.$$
 
$$-\sin(y)y'=1.$$
 
If we write $\theta=\mathrm{arccos}(z)$ then the following image shows that $\cos(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br />
 
If we write $\theta=\mathrm{arccos}(z)$ then the following image shows that $\cos(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br />
[[File:Arccosderivativetriangle.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 />
 
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
 
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$

Revision as of 05:01, 28 October 2014

The function $\mathrm{arccos} \colon [-1,1] \longrightarrow [0,\pi]$ is the inverse function of the cosine function.

Arccos.png

Complex arccos.jpg

Properties

Proposition: $$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}}$$

Proof: If $y=\mathrm{arccos}(z)$ then $\cos(y)=z$. Now use implicit differentiation with respect to $z$ to get $$-\sin(y)y'=1.$$ If we write $\theta=\mathrm{arccos}(z)$ then the following image shows that $\cos(\mathrm{arccos}(z))=\sqrt{1-z^2}$:

Sin(arccos(z)).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:

References

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