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?

## Trig Problem

**Moderators:** gmalivuk, Moderators General, Prelates

### 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();}

### 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.

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.

### 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();}

### 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...

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...

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