Difference between revisions of "Arcsin"
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| − | The $\mathrm{arcsin}$ | + | The function $\mathrm{arcsin} \colon [-1,1] \rightarrow \left[ \dfrac{-\pi}{2}, \dfrac{\pi}{2} \right]$ is the [[inverse function]] of the [[sine]] function. <br /> |
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| − | [ | + | <div align="center"> |
| + | <gallery> | ||
| + | File:Arcsin.png|Graph of $\mathrm{arcsin}$ on $[-1,1]$. | ||
| + | File:Complex arcsin.jpg|[[Domain coloring]] of [[analytic continuation]]. | ||
| + | </gallery> | ||
| + | </div> | ||
=Properties= | =Properties= | ||
Revision as of 04:32, 31 October 2014
The function $\mathrm{arcsin} \colon [-1,1] \rightarrow \left[ \dfrac{-\pi}{2}, \dfrac{\pi}{2} \right]$ is the inverse function of the sine function.
- Arcsin.png
Graph of $\mathrm{arcsin}$ on $[-1,1]$.
- Complex arcsin.jpg
Properties
Proposition: $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}}$$
Proof: If $y=\mathrm{arcsin}(z)$ then $\sin(y)=z$. Now use implicit differentiation with respect to $z$ to get $$\cos(y)y'=1.$$ Substitution back in $y=\mathrm{arcsin(z)}$ 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: █
