[SOLVED] Trig Functions

**>_<SHY_GUY>_<** · February 15th 2009, 11:50 AM

This Is Really Bad, Ive Frgot What To Do With The Problem Below:

y= 3csc (3x+pi) - 2

it asks for :
1. the period
2. domain
3. range
4 the graph...but there is like something that helps to clue you in graphing, i dont remember...im reading through the book.

this is what i did with the equation:

3csc [3(x + pi/3)] - 2
where do i go after that?

**Moo** · February 15th 2009, 11:54 AM

Buenos dias

Originally Posted by >_<SHY_GUY>_<

This Is Really Bad, Ive Frgot What To Do With The Problem Below:

y= 3csc (3x+pi) - 2

it asks for :
1. the period
2. domain
3. range
4 the graph...but there is like something that helps to clue you in graphing, i dont remember...im reading through the book.

this is what i did with the equation:

3csc [3(x + pi/3)] - 2
where do i go after that?

For the period, let t be the period, t>0.
f(x+t)=f(x)
f(x+t)=3 csc (3x+3t+pi)-2
but you know that csc has a period 2pi, that is to say csc(z+2pi)=csc(z)
so 3t=2pi
--> t=2pi/3

the domain is the set of all real numbers, except the ones for which csc is undefined (which are the ones that annulate a sine) :
solve for x in 3x+pi=k*pi, for any (positive or negative) integer k.

do these help ?

**>_<SHY_GUY>_<** · February 15th 2009, 12:01 PM

Originally Posted by Moo

Buenos dias

For the period, let t be the period, t>0.
f(x+t)=f(x)
f(x+t)=3 csc (3x+3t+pi)-2
but you know that csc has a period 2pi, that is to say csc(z+2pi)=csc(z)
so 3t=2pi
--> t=2pi/3

the domain is the set of all real numbers, except the ones for which csc is undefined (which are the ones that annulate a sine) :
solve for x in 3x+pi=k*pi, for any (positive or negative) integer k.

do these help ?

Ah! Bonour Mademoiselle "Moo" =]
Comment Allez-Vous?

ok, i got how to determine the period....Muchas Gracias

but the only thing that really confuses me is finding domain and ranges of graph, especially for trig functions :/

**Moo** · February 15th 2009, 12:06 PM

Originally Posted by >_<SHY_GUY>_<

Ah! Bonour Mademoiselle "Moo" =]
Comment Allez-Vous?

ok, i got how to determine the period....Muchas Gracias

but the only thing that really confuses me is finding domain and ranges of graph, especially for trig functions :/

ça va, merci

$\csc(z)=\frac{1}{\sin(z)}$
So it's not defined if $\sin(z)=0$ .
$\sin(z)=0 \Leftrightarrow z=k \pi ~,~ k \in \mathbb{Z}$ (the set of all integers)

So for $\csc(3x+\pi)$ to be defined, you need $3x+\pi$ to be different from $k \pi$
This means $3x+\pi \neq k \pi \Rightarrow 3x \neq k \pi$ (since k is any integer)
So it's not defined for $x=\frac{k\pi}{3}$ , for any integer k.

The domain is then $D=\mathbb{R}- \left\{\tfrac{k \pi}{3} ~:~ k \in \mathbb{Z}\right\}$

Does it look clear ?

For the range... the cosecant's range is $\mathbb{R}-(-1,1)$ (see here : http://mathworld.wolfram.com/Cosecant.html )
So for any z in the domain of the cosecant, csc(z) $\geq$ 1 or csc(z) $\leq$ -1. Thus 3csc(z) $\geq$ 3 or 3csc(z) $\leq$ -3. --> 3csc(z)-2 $\geq$ 1 or 3csc(z) $\leq$ -5.

The range of $3 \csc(3x+\pi)-2$ is hence $\mathbb{R}-(-5,1)$

For the graph, I have no particular method... I'm sorry, because we never learnt csc so I can't help you more than your book on this :s

**>_<SHY_GUY>_<** · February 15th 2009, 12:17 PM

Originally Posted by Moo

ça va, merci

$\csc(z)=\frac{1}{\sin(z)}$
So it's not defined if $\sin(z)=0$ .
$\sin(z)=0 \Leftrightarrow z=k \pi ~,~ k \in \mathbb{Z}$ (the set of all integers)

So for $\csc(3x+\pi)$ to be defined, you need $3x+\pi$ to be different from $k \pi$
This means $3x+\pi \neq k \pi \Rightarrow 3x \neq k \pi$ (since k is any integer)
So it's not defined for $x=\frac{k\pi}{3}$ , for any integer k.

The domain is then $D=\mathbb{R}- \left\{\tfrac{k \pi}{3} ~:~ k \in \mathbb{Z}\right\}$

Does it look clear ?

For the range... the cosecant's range is $\mathbb{R}-(-1,1)$ (see here : Cosecant -- from Wolfram MathWorld )
So for any z in the domain of the cosecant, csc(z) $\geq$ 1 or csc(z) $\leq$ -1. Thus 3csc(z) $\geq$ 3 or 3csc(z) $\leq$ -3. --> 3csc(z)-2 $\geq$ 1 or 3csc(z) $\leq$ -5.

The range of $3 \csc(3x+\pi)-2$ is hence $\mathbb{R}-(-5,1)$

For the graph, I have no particular method... I'm sorry, because we never learnt csc so I can't help you much than your book on this :s

wow you are good...i think you learn it in a more complex [i think] method...but for range i think i got it ^_^

for domain, i see that csc is undefined every xpi , and we have 3x + pi... but i would think to put them equal to each other, but it doesnt work out...would i just have to ignore the pi, or is that just wrong?

muchas gracias por haberme ayudado =]

**Moo** · February 15th 2009, 12:20 PM

wow you are good...i think you learn it in a more complex [i think] method...but for range i think i got it ^_^

Well, it was not a particular complex method, I just thought it would be clearer to see it this way...
No sé como tu has aprendido :P

for domain, i see that csc is undefined every xpi , and we have 3x + pi... but i would think to put them equal to each other, but it doesnt work out...would i just have to ignore the pi, or is that just wrong?

Yes, you can ignore pi.
Let 3x+pi=k pi (not xpi, it can be confusing because you already have one)
It's equivalent to 3x=k pi, because k is any integer, so it doesn't matter whether you consider k or k-1.

muchas gracias por haberme ayudado =]

de nada ?

**>_<SHY_GUY>_<** · February 15th 2009, 12:27 PM

Originally Posted by Moo

Well, it was not a particular complex method, I just thought it would be clearer to see it this way...
No sé como tu has aprendido :P

Yes, you can ignore pi.
Let 3x+pi=k pi (not xpi, it can be confusing because you already have one)
It's equivalent to 3x=k pi, because k is any integer, so it doesn't matter whether you consider k or k-1.

de nada ?

Pues La manera que me estas ensenando es mas diferente de lo que nos ensenan en clase

ok, now i get it...i mean i get blurry up to a point, but once i get past it, everything comes back to me quick...

Muchisimas Gracias Senorita

Haber Cuando Nos Hablamos eh?

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