What is the difference between y=asin(sin(x)) and y=sin(asin(x))? (1 Viewer)

sinophile · Jun 8, 2009

And why are their domains different? Because I thought they were inverse functions and cancelled each other out to become the same function y=x. Can someone explain?

study-freak · Jun 8, 2009

sinophile said:
And why are their domains different? Because I thought they were inverse functions and cancelled each other out to become the same function y=x. Can someone explain?

do you mean y=sin(arcsinx)?

Fortify · Jun 8, 2009

study-freak said:
do you mean y=sin(arcsinx)?

asinx is the same as sin^-1 x

asinx is used in Graphmatica and GeoGebra.

study-freak · Jun 8, 2009

$\\y=sin^{-1}(sinx)=sin^{-1}f(x)$ which has domain restriction $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2} \\y=sin(sin^{-1}x)=sinf(x)$ which does not have domain restriction.$$
is what I think.

Dumbledore · Jun 8, 2009

study-freak said:
$\\y=sin^{-1}(sinx)=sin^{-1}f(x)$ which has domain restriction $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2} \\y=sin(sin^{-1}x)=sinf(x)$ which does not have domain restriction.$$
is what I think.

i think u mixed it up a bit
sin(asin(x)) will have a domain of -1-x-1 since the domain of asin(x) is -1-x-1.

and asin(sin(x)) has infinite domain because the sin of any value of x will lie between -1-x-1 so the asin of the number will always have a value

can u use greater than signs in this? it seems to cut off my text when i use it <>

tommykins · Jun 8, 2009

y = asin(sinx) has no restricted domain, as sin x oscillates between -1 <= x <= 1 and is valid within the domain of the function asin

y = sin(asinx) has a restricted domain of -1 < x < 1 as asin is only valid within those regions.

Dumbledore · Jun 8, 2009

also the asin(sin(x)) do not cancel completely to x the graph of asin(sin(x)) is given by
$sin^{-1}(sin\: x)=\int \frac{d}{dx}[sin^{-1}(sin\: x)]dx\\ u = sin\: x;\:\; \: y = sin^{-1}u\\ \frac{du}{dx}=cos\: x;\; \frac{dy}{du}= \frac{1}{\sqrt{1-u^{2}}}\\ \frac{d}{dx}[sin^{-1}(sin\: x)]= \frac{cos\: x}{\sqrt{1-sin^{2}x}}\\ =\frac{cos\: x}{\sqrt{cos^{2}x}} \\=\frac{cos\: x}{|cos\: x|}\\sin^{-1}(sin\: x)= \\=x\: given\: cos\: x>0 \: or \: -\frac{\pi}{2}+2n\pi<x<\frac{\pi}{2}+2n\pi \\=-x\: given\: cos\: x>0 \: or \: \frac{\pi}{2}+2n\pi<x<\frac{3\pi}{2}+2n\pi$
when u use latex do u memorise the syntax and type it in or just use the buttons?

Lebosouljah · Jun 8, 2009

difference is y = x^2 sin 23

tommykins · Jun 8, 2009

Dumbledore said:
also the asin(sin(x)) do not cancel completely to x the graph of asin(sin(x)) is given by
$sin^{-1}(sin\: x)=\int \frac{d}{dx}[sin^{-1}(sin\: x)]dx\\ u = sin\: x;\:\; \: y = sin^{-1}u\\ \frac{du}{dx}=cos\: x;\; \frac{dy}{du}= \frac{1}{\sqrt{1-u^{2}}}\\ \frac{d}{dx}[sin^{-1}(sin\: x)]= \frac{cos\: x}{\sqrt{1-sin^{2}x}}\\ =\frac{cos\: x}{\sqrt{cos^{2}x}} \\=\frac{cos\: x}{|cos\: x|}\\sin^{-1}(sin\: x)= \\=x\: given\: cos\: x>0 \: or \: -\frac{\pi}{2}+2n\pi<x><\frac{\pi}{2}+2n\pi \\=-x\: given\: cos\: x>0 \: or \: \frac{\pi}{2}+2n\pi<x><\frac{3\pi}{2}+2n\pi$
when u use latex do u memorise the syntax and type it in or just use the buttons?

i remember the syntax
</x></x>

study-freak · Jun 9, 2009

Dumbledore said:
i think u mixed it up a bit
sin(asin(x)) will have a domain of -1-x-1 since the domain of asin(x) is -1-x-1.

and asin(sin(x)) has infinite domain because the sin of any value of x will lie between -1-x-1 so the asin of the number will always have a value

can u use greater than signs in this? it seems to cut off my text when i use it <>

lol i did mix it up..

Pwnage101 · Jun 9, 2009

graph them

also graph cos(acosx) and tan(atanx) and acso(cosx) and atan(tanx)

Timothy.Siu · Jun 11, 2009

they're probably zig zag lines lol

What is the difference between y=asin(sin(x)) and y=sin(asin(x))? (1 Viewer)

sinophile

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