Difference between revisions of "Arccos"

From specialfunctionswiki
Jump to: navigation, search
(Properties)
(Properties)
Line 13: Line 13:
 
$$-\sin(y)y'=1.$$
 
$$-\sin(y)y'=1.$$
 
Substituting back in $y=\mathrm{arccos}(z)$ yields the formula
 
Substituting back in $y=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.$$
+
$$\dfrac{d}{dz} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.$$   
 
 
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 23:08, 25 October 2014

The $\mathrm{arccos}$ function 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 $$-\sin(y)y'=1.$$ 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