Math help !

- anonymous

Math help !

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- anonymous

@terenzreignz

- anonymous

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- terenzreignz

Domain of validity would simply be the domain of the csc function

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- anonymous

Alrighty, What's next?

- terenzreignz

That's it. Just find the domain of the csc function, and that'd be your answer.

- anonymous

There a set of real numbers right ?

- terenzreignz

All of it? No. There are certain real numbers that are not in the domain of csc.

- anonymous

Is it a set of real number except for multiples of pi

- terenzreignz

That's much better :)

- terenzreignz

Because multiples of pi make sin(x) equal to zero, meaning 1/sin(x) would be 1/0, which can't be allowed.

- anonymous

So this phrase "Is it a set of real number except for multiples of pi" would be my answer ?

- terenzreignz

Of course not, that's a question. Be more confident...
"The set of real numbers except for multiples of pi"
LOL
or, in notation, it would be
\[\LARGE \mathbb{R} \backslash \left\{n\pi \left| \quad n \in \mathbb{Z}\right.\right\}\]

- anonymous

Simplify trigonometric expression

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- terenzreignz

Okay, I need you to recall that...
\[\LARGE x^2-y^2 = (x+y)(x-y)\]
remember? :)

- anonymous

Yes, remember.

- terenzreignz

Okay, great... now, can you express \(\large \sin^2 \theta\) in terms of cos?

- anonymous

Correct. One bad news. My calulator died & i don't have batteries for it lol

- terenzreignz

There is no need for calculators (hopefully) and you have google.
Anyway, if you do know, then express \(\large \sin^2 \theta \) in terms of cos.

- anonymous

yes correct.

- terenzreignz

Correct?? I just asked you a question :/

- anonymous

Sorry i didn't write that. what do you mean by the express sin^2 theta of cos ?

- terenzreignz

I mean just that... express sin^2 in terms of cos... Recall the pythagorean identity
\[\Large \sin^2 \theta + \cos^2 \theta = 1\]
And just solve for \(\large \sin^2 \theta\)

- anonymous

Oh okay so would my answer be 1+sin (theta) over cos(theta) ?

- terenzreignz

Don't jump to conclusions, we will do this step-by-step.
So... solving for \(\large \sin^2 \theta\) from the pythagorean identity, what do you get?

- anonymous

I'm a little confused how could i solve that

- terenzreignz

It's simple... using algebraic manipulation, just bring everything that is not \(\large \sin^2\theta\) to the other side.

- anonymous

I just have to work on this problem because i don't really understand .

- terenzreignz

All I really needed you to do was
\[\Large \sin^2 \theta + \cos^2 \theta = 1\]
subtract \(\large \cos^2 \theta\) from both sides...
\[\Large \sin^2 \theta + \cos^2 \theta \color{red}{-\cos^2 \theta} = 1\color{red}{-\cos^2 \theta}\]

- terenzreignz

Simplifying, we get...
\[\Large \sin^2\theta = 1- \cos^2 \theta\]

- terenzreignz

And that's how to express \(\large \sin^2 \theta\) in terms of cos.

- anonymous

Ohhhhhhhhh okay now i see. So after we do that what do we do to get the answer ?

- terenzreignz

Take a good look at the right-side of the equation... does it look familiar to you?
\[\Large \sin^2\theta = \color{blue}{1- \cos^2 \theta}\]
I asked you to recall that
\[\large x^2 - y^2 = (x+y)(x-y)\]
Something similar can be done to the right-side of the equation...

- anonymous

with that 1-cos^2(theta) will it be over sin(theta)

- terenzreignz

Don't get ahead of yourself... using what I asked you to recall
(namely that \(x^2 - y^2 = (x+y)(x-y)\) )
Factor the right side... of this equation
\[\Large \sin^2\theta = \color{blue}{1- \cos^2 \theta}\]

- anonymous

I'm still not understanding

- terenzreignz

maybe a simple tweak?
\[\large 1 = 1^2\]
\[\Large \sin^2 \theta = 1-\cos^2 \theta = \color{blue}{1^2 -\cos^2 \theta}\]

- anonymous

ok

- terenzreignz

This pattern... use it...\[\huge x^\color{red}{2}-y^\color{red}2= (x+y)(x-y)\]

- anonymous

okay i understand that by looking at it

- terenzreignz

Well, you have to learn to recognise it.. it's here...
\[\Large \sin^2 \theta= \color{blue}{1^2 - \cos^2 \theta}\]

- terenzreignz

The pattern is called the difference of two squares.

- anonymous

Okay got that

- terenzreignz

So...?

- anonymous

Do i solve that above

- terenzreignz

To be blunt...
\[\large x^2 - y^2 \] factors into
\[\large (x+y)(x-y)\]
so to what does
\[\Large 1^2 - \cos^2 \theta\]factor into?

- anonymous

1^2 + cos^2(theta)

- terenzreignz

I suggest you review factoring "difference of two squares"...

- anonymous

I am going to review after we finish this problem. I know your trying to go to bed now

- terenzreignz

Yes... yes you do :)
but how can we finish this problem without knowing how to factor a difference of two squares? :)

- anonymous

I can just re view after this problem lol

- terenzreignz

Very well.. then back to my query...
to what does
\[\Large 1^2 - \cos^2 \theta\] factor into?

- anonymous

I thought i just answered this 3 times lol

- terenzreignz

Three times incorrectly. Reread through what I asked you to recall before we even started working on this problem, the answer is there...

- anonymous

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- terenzreignz

Now, was that really necessary? :D

- anonymous

Haha no i just want you to go to sleep

- terenzreignz

Oh, trust me, I do want to go to sleep... :)
but I don't like leaving things undone...
I'll finish this or die trying... so to speak...
Now, my request? That factoring thing?

- anonymous

Are we back to the factoring thing. I just need to finish this one i have 20 more problems to go.

- terenzreignz

Well, then, let's finish it :)
Starting with factoring that expression there...

- anonymous

Okay is there an easy way to do this because im conused. You can just go to bed ill just skip this problem

- terenzreignz

Is that giving up?

- anonymous

Well no lol i just dn't understand

- terenzreignz

Well, while you're at it, review factoring a difference of two squares... it comes in handy in trig.

- anonymous

i am going to review .

- terenzreignz

Good. That's all I want to know :)
-------------------------------------
Terence out

- anonymous

Whats the answer ?

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