anonymous
  • anonymous
Math help !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@terenzreignz
anonymous
  • anonymous
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terenzreignz
  • terenzreignz
Domain of validity would simply be the domain of the csc function

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anonymous
  • anonymous
Alrighty, What's next?
terenzreignz
  • terenzreignz
That's it. Just find the domain of the csc function, and that'd be your answer.
anonymous
  • anonymous
There a set of real numbers right ?
terenzreignz
  • terenzreignz
All of it? No. There are certain real numbers that are not in the domain of csc.
anonymous
  • anonymous
Is it a set of real number except for multiples of pi
terenzreignz
  • terenzreignz
That's much better :)
terenzreignz
  • 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
  • anonymous
So this phrase "Is it a set of real number except for multiples of pi" would be my answer ?
terenzreignz
  • 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
  • anonymous
Simplify trigonometric expression
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terenzreignz
  • terenzreignz
Okay, I need you to recall that... \[\LARGE x^2-y^2 = (x+y)(x-y)\] remember? :)
anonymous
  • anonymous
Yes, remember.
terenzreignz
  • terenzreignz
Okay, great... now, can you express \(\large \sin^2 \theta\) in terms of cos?
anonymous
  • anonymous
Correct. One bad news. My calulator died & i don't have batteries for it lol
terenzreignz
  • 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
  • anonymous
yes correct.
terenzreignz
  • terenzreignz
Correct?? I just asked you a question :/
anonymous
  • anonymous
Sorry i didn't write that. what do you mean by the express sin^2 theta of cos ?
terenzreignz
  • 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
  • anonymous
Oh okay so would my answer be 1+sin (theta) over cos(theta) ?
terenzreignz
  • 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
  • anonymous
I'm a little confused how could i solve that
terenzreignz
  • terenzreignz
It's simple... using algebraic manipulation, just bring everything that is not \(\large \sin^2\theta\) to the other side.
anonymous
  • anonymous
I just have to work on this problem because i don't really understand .
terenzreignz
  • 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
  • terenzreignz
Simplifying, we get... \[\Large \sin^2\theta = 1- \cos^2 \theta\]
terenzreignz
  • terenzreignz
And that's how to express \(\large \sin^2 \theta\) in terms of cos.
anonymous
  • anonymous
Ohhhhhhhhh okay now i see. So after we do that what do we do to get the answer ?
terenzreignz
  • 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
  • anonymous
with that 1-cos^2(theta) will it be over sin(theta)
terenzreignz
  • 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
  • anonymous
I'm still not understanding
terenzreignz
  • 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
  • anonymous
ok
terenzreignz
  • terenzreignz
This pattern... use it...\[\huge x^\color{red}{2}-y^\color{red}2= (x+y)(x-y)\]
anonymous
  • anonymous
okay i understand that by looking at it
terenzreignz
  • terenzreignz
Well, you have to learn to recognise it.. it's here... \[\Large \sin^2 \theta= \color{blue}{1^2 - \cos^2 \theta}\]
terenzreignz
  • terenzreignz
The pattern is called the difference of two squares.
anonymous
  • anonymous
Okay got that
terenzreignz
  • terenzreignz
So...?
anonymous
  • anonymous
Do i solve that above
terenzreignz
  • 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
  • anonymous
1^2 + cos^2(theta)
terenzreignz
  • terenzreignz
I suggest you review factoring "difference of two squares"...
anonymous
  • anonymous
I am going to review after we finish this problem. I know your trying to go to bed now
terenzreignz
  • terenzreignz
Yes... yes you do :) but how can we finish this problem without knowing how to factor a difference of two squares? :)
anonymous
  • anonymous
I can just re view after this problem lol
terenzreignz
  • terenzreignz
Very well.. then back to my query... to what does \[\Large 1^2 - \cos^2 \theta\] factor into?
anonymous
  • anonymous
I thought i just answered this 3 times lol
terenzreignz
  • 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
  • anonymous
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terenzreignz
  • terenzreignz
Now, was that really necessary? :D
anonymous
  • anonymous
Haha no i just want you to go to sleep
terenzreignz
  • 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
  • anonymous
Are we back to the factoring thing. I just need to finish this one i have 20 more problems to go.
terenzreignz
  • terenzreignz
Well, then, let's finish it :) Starting with factoring that expression there...
anonymous
  • 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
  • terenzreignz
Is that giving up?
anonymous
  • anonymous
Well no lol i just dn't understand
terenzreignz
  • terenzreignz
Well, while you're at it, review factoring a difference of two squares... it comes in handy in trig.
anonymous
  • anonymous
i am going to review .
terenzreignz
  • terenzreignz
Good. That's all I want to know :) ------------------------------------- Terence out
anonymous
  • anonymous
Whats the answer ?

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