## Arccosine (trig function questions)

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brain_ofj
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### Inverse Trigonometric Functions questions...

It's been awhile and I'm having trouble finding resources about inverse trig functions... plenty on the basic trig functions...

Ok, so $cos^(-1)$ is arccosine.

Which means I can treat it just like any other function in regards to division/ multiplication to move to other sides of an equal sign... right?

For instance,

G(x) = arccos(f(x))

cos*G(x) = f(x) ?

But this would mean that $sec\theta = 1/ cos\theta = arccos$...

But that does seem right to me...

I'm trying to find a simpler form of this function

G(x) = arccos(l *sin(2x))

Any help would be appreciated.

Timefly
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### Re: Inverse Trigonometric Functions questions...

G(x) = acos(f(x))
cos(G(x)) = f(x)

Agus
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### Re: Inverse Trigonometric Functions questions...

Actually, f * f-1 isn't the same as f(f-1)..

You are messing up with the inverse of a number and the inverse of a function.
A function (x) applied to the inverse of the function does give you (x).
A function multiplied by the inverse of the function gives you a new function..

Dopefish
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### Re: Inverse Trigonometric Functions questions...

arccos(x) is sometimes written as cos-1(x), but that is different from cos(x)-1, which seems to be the source of your confusion.

cos-1(x) is the inverse function of cos (so it in effect undoes cos), where as cos(x)-1 is the reciprocal of cos(x) aka sec(x), which is definitely something different.

iamacow
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### Re: Arccosine (trig function questions)

Simply put, the problem is that f-1(x) isn't the same as f(x)-1
so arccos(x) = cos-1(x) = the function such that cos(arccos(x)) = arccos(cos(x)) = x (on the right domain)
and sec(x) = cos(x)-1 = 1/cos(x) the function such that cos(x)*sec(x) = 1 (on the right domain)

EDIT: I should probably say that they usually aren't the same. It's confusing because for trig functions people like to write stuff like sin2(x) for the "square of the sin of x" even though that's confusing and not usually (in my experience) what people mean for other functions when they write fn(x)

brain_ofj
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### Re: Arccosine (trig function questions)

Ok. Thank you.
That makes sense.

Now, if I want to "take the arccosine" to the left hand side of the equation... how would I do that?
I can't divide/ multiply through like I would using normal arithmetic...

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### Re: Arccosine (trig function questions)

You apply cos to both sides. The cos and the arccos cancel out, leaving you with a cos on the left, which is correct. (A simpler example is x=y/2 into 2x=y by multiplying both sides by 2.)
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brain_ofj
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### Re: Arccosine (trig function questions)

Alright... duh.

Making it cos(G(x)) = l*sin(2x), right?
Last edited by brain_ofj on Wed Apr 18, 2012 2:27 am UTC, edited 1 time in total.

Sagekilla
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### Re: Arccosine (trig function questions)

brain_ofj wrote:Alright... duh.

Making it cos(G(x)) = l*sin(2x)

This is one of those instances where, despite how nice the short hand is, I like to write out the inverse
trig functions as arc* (arctan, arcsin, etc). I actually made the same mistake you did when I took my calc
1 class. We had to do implicit differentiation and I accidentally treated an inverse sine (written as sin to -1)
as 1 / sin.
http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.

brain_ofj
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Joined: Thu Feb 09, 2012 3:32 am UTC

### Re: Arccosine (trig function questions)

Alright, since this is a function though, is the above correct?
ie, cos(G(x)) = l*sin(2x) ?

jestingrabbit
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### Re: Arccosine (trig function questions)

Yes, that's right.
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gfauxpas
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### Re: Arccosine (trig function questions)

Another thing to watch out for, the -1 in general implies the inverse on the entire domain, f: X → Y, f-1: Y → X. But none of the trig functions are invertible without restricting the domain. So sin-1 can imply a relation that's one-to-many, but arcsine is something more specific than "the inverse of sine".

fishfry
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### Re: Inverse Trigonometric Functions questions...

brain_ofj wrote:It's been awhile and I'm having trouble finding resources about inverse trig functions... plenty on the basic trig functions...

Ok, so $cos^(-1)$ is arccosine.

Which means I can treat it just like any other function in regards to division/ multiplication to move to other sides of an equal sign... right?

No, you're getting fooled by a notational ambiguity.

The arccos is the inverse function of the cos. It's not the multiplicative reciprocal. So when we write cos^(-1)(x) we do NOT mean 1 over the cosine of x. We mean the angle whose cosine is x.

The notation is perhaps not very good, because generations of students get tripped up by it. But you need to be aware that cos^(-1) is the inverse function, not the multiplicative reciprocal.