Difference between revisions of "Arcsin"
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$$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{1-z^2}$$ | $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{1-z^2}$$ | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> 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 | ||
| + | $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}}. █ $$ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 01:46, 28 October 2014
The $\mathrm{arcsin}$ function is the inverse function of the sine function.
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Properties
Proposition: $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{1-z^2}$$
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: █
