F28

Arcusfunktioner

Inverser till sinus, cosinus, tangens mm.

Arcsin

Inversen till $y=\sin x$ $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
kallas arcsinus (${\sin }^{-1}x$)
$y=\arcsin x$
$\iff$
$x=\sin y$, $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

Kanceleringslagar
$\arcsin (\sin (x))=x$, $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
$\sin (\arcsin (x))=x$, $-1\le x\le 1$

Derivatan

$y=\arcsin (x)\iff \sin y=x$
$\frac{d}{dx}\sin y=\frac{d}{dx}(x)$
$\cos (y)-y^{\prime }=1$
$y^{\prime }=\frac{1}{\cos y}$, $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$

$\cos y=\sqrt{1^2-x^2}=\sqrt{1-x^2}$

$\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}$

Arctan

$\tan (x)$, $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
$\frac{d}{dx}(\tan (x))=\frac{1}{{\cos }^{2}x}=(1+{\tan }^{2}x)>0$ på $(-\frac{\pi }{2},\frac{\pi }{2})$

$\tan x$ är 1-1 Inversen kallas arctangens
$y=\arctan (x)\iff \tan (y)=x$, $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$
$\arctan (\tan (x))=x$, $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
$\tan (\arctan (x))=x$, $-\infty <x<\infty$

Derivatan

$y=\arctan (x)\iff \tan y=x$
Implicit derivering $\frac{d}{dx}(\tan y)=\frac{d}{dx}(x)$
$(1+{\tan }^{2}y)y^{\prime }=1$

$\tan (y)=\frac{x}{1}=x$
$1+{\tan }^2(y)=1+x^2$
$y^{\prime }=\frac{1}{1+{\tan }^2(y)}$
$y^{\prime }=\frac{1}{1+x^2}$

Arccos

$\cos (y)=\sin (\frac{\pi }{2}-y)$, $0\le y\le \pi$

Inversen av $\cos (x)$, $0\le y\le \pi$
kallas arcuscosinus ($\arccos (x)$)

$y=\arccos (x)\iff \cos y=x$, $0\le y\le \pi$
$\sin (\frac{\pi }{2}-y)=x$
$\iff$
$\frac{\pi }{2}-y=\arcsin x$
$y=\arccos x$ ger att
$\frac{\pi }{2}-\arccos x=\arcsin x$
$\iff$
$\arccos x=\frac{\pi }{2}-\arcsin x$, $-1\le x\le 1$

Derivatan

$\frac{d}{dx}(\arccos x)=\frac{-1}{\sqrt{1-x^2}}$

$\frac{d}{dx}\arccos (x)=\frac{d}{dx}(\frac{\pi }{2})-\frac{d}{dx}(\arcsin x)$
$=0-\frac{1}{\sqrt{1-x^2}}=-\frac{1}{1-x^2}$