Difference between revisions of "Arcsin"

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=References=
 
=References=
 
*[http://mathworld.wolfram.com/InverseSine.html  Weisstein, Eric W. "Inverse Sine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseSine.html]
 
*[http://mathworld.wolfram.com/InverseSine.html  Weisstein, Eric W. "Inverse Sine." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseSine.html]

Revision as of 05:57, 31 October 2014

The function $\mathrm{arcsin} \colon [-1,1] \rightarrow \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ is the inverse function of the sine function.

Properties

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

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

Cos(arcsin(z)).png

Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}}. █$$

Proposition: $\int \mathrm{arcsin}(z) dz = \sqrt{1-z^2}+z\mathrm{arcsin}(z)+C$

Proof:

Proposition: $\mathrm{arcsin}(z) = \mathrm{arccsc}\left( \dfrac{1}{z} \right)$

Proof:

Proposition: $\mathrm{arcsin}(z)=\sum_{k=0}^{\infty} \dfrac{\left(\frac{1}{2} \right)_n}{(2n+1)n!}x^{2n+1}$

Proof:


References