Trig Problem

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Plarin
Posts: 27
Joined: Sat Apr 25, 2009 8:58 pm UTC

Trig Problem

This is a problem I've been wondering about for a couple of days now. I showed my Calc teacher, but he couldn't help, so here it is:

1. x = 135°
2. sin x = √(2)/2
3. arcsin(sin x) = arcsin((√2)/2)
4. x = 45°
5. 135° = 45°

I wanna say that step 3 is incorrect, but then that would mean that f(x) =/= f(x). So, any help?

Qoppa
Posts: 694
Joined: Sat Nov 24, 2007 9:32 pm UTC
Location: Yes.

Re: Trig Problem

The inverse trig functions only return the principle value of their input.

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

Plarin
Posts: 27
Joined: Sat Apr 25, 2009 8:58 pm UTC

Re: Trig Problem

What do you mean by principal value? There's only one output paired with each input.

Also, are you saying that f(x) is not always equal to f(x)?

Edit- Oh wait, you're saying arcsin(sin x) =/= x. That makes sense. I guess I just overlooked that part out of sheer habit.

Qoppa
Posts: 694
Joined: Sat Nov 24, 2007 9:32 pm UTC
Location: Yes.

Re: Trig Problem

Exactly. [imath]\sin(\pi/4)=\sin(3\pi/4)=\sin(9\pi/4)=\ldots=\sqrt{2}/2[/imath], but for the inverse to be a function, you can't have [imath]\arcsin(\sqrt(2)/2)=\pi/4, 3\pi/4\ldots.[/imath], so we take the principle value, which for arcsin, will give you whichever value works in the interval [imath][-\pi/2, \pi/2][/imath].

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

mr-mitch
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

Re: Trig Problem

sine is a periodic function, it doesn't really have an inverse for the domain of R (many to one function becomes one-to-many not a function), so arcsine is limited to the range [-pi/2,pi/2] as stated by the previous post.

arccosine is limited to [0,pi], and arctangent is limited to [-pi/2,pi/2].

Effectively they return the first value increasing from -pi/2 which satisfies the sine equation...