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TRIG HELP

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Bodhisattva
Posts: 878 - Karma: 138
Tommorow I have a test for Trig but my teacher gave us one problem to do on our own as homework and bring it in for credit on the test. Can anyone prove this problem and explain how they did it?This is the problem:
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Oct 26 2010 8:11PM
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altadonkeys
Posts: 122 - Karma: 118
I can't remember how to do any of those god awful trig proofs.
Oct 26 2010 8:13PM
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teleak
Posts: 2251 - Karma: 105
SohCahToa
Oct 26 2010 8:17PM
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AbeFroman
Posts: 1268 - Karma: -4
i think if anyone can read that they should get extra credit
Oct 26 2010 8:19PM
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altadonkeys
Posts: 122 - Karma: 118
www.wolframalpha.com
it does trigonometry and shows the work
Oct 26 2010 8:27PM
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Bodhisattva
Posts: 878 - Karma: 138
I wrote it in Paint really quickly with my laptop touch pad ahha
Oct 26 2010 9:31PM
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2_fadez
Posts: 7780 - Karma: 4,183
I'm going to do this for you because it's starting to get on my nerves that no one has done it yet.
Oct 26 2010 9:34PM
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Gingeski
Posts: 100 - Karma: 16
I'm trying it and it is not going so well. . .
Oct 26 2010 9:36PM
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p-fo
Posts: 9372 - Karma: 6,155
fuck this one is hard and im in calc!
Oct 26 2010 9:41PM
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2_fadez
Posts: 7780 - Karma: 4,183
this fucker took me literally 60 seconds.
uploading pics is the hard part

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Oct 26 2010 9:42PM
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Bodhisattva
Posts: 878 - Karma: 138
I get everything you did up until the last 2 steps. It looks like you switched cot^2 x/cos x to the other side? You can't do that can you?
Oct 26 2010 10:16PM
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BootySpunk.
Posts: 8148 - Karma: 84
What the fuck kind of voodoo shit did you pull there?
Oct 26 2010 10:18PM
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Schlop
Posts: 3640 - Karma: 24
IT MEANS "MATH IS AWESOME" IN JAPANESE!
Oct 26 2010 10:19PM
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2_fadez
Posts: 7780 - Karma: 4,183
Just so you know, I switched cot^2 into cos^2/sin^2 pretty early on, so I'm not sure what you are referring to
Oct 26 2010 10:21PM
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ARH
Posts: 117 - Karma: 14
is there any given about csc^2-1 equalling something?  there prolly is. 
Oct 26 2010 10:23PM
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Bodhisattva
Posts: 878 - Karma: 138
I mean cos^2 x/sin^2 x
Oct 26 2010 10:23PM
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BootySpunk.
Posts: 8148 - Karma: 84
It is equal to cot^2 and than you have to separate that into cos^2/sin^2 and cancel out one of the cos...I am still trying to figure out how to deal with the 1/1+cosx
Oct 26 2010 10:25PM
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Gingeski
Posts: 100 - Karma: 16
You can, but you don't have to. Sorry for lack of pictures, but instead of moving it to the other side, you can just give it a common denominator. So multiply the 1/(1+cosx) by sin^2x/sin^2x and then multiply the cosx/sin^2x by 1+cosx/1+cosx. Then combined youll have sin^2x+cosx+cos^2x all over sin^2x(1+cosx). In the numerator, the sin^2x and cos^2x will add to one (original identity. So you're left with 1+cosx over sin^2x(1+cosx), the 1+cosx's will cancel out leaving you with 1/sin^2x which is equal to csc^2x. Hopefully you can follow all this mumbo jumbo
Oct 26 2010 10:27PM
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123+ski
Posts: 2832 - Karma: 581
cross multiply on the last step and you get 1-cos^2x=sin^2x, which you know is true because sin^2x+cos^2x=1
i didn't really get what he was doing in the right corner, specifically the part where 1/(1+cosx) turns into 1, but other than that its legit. so follow his method until the last line, then add my step before u get to the right corner, and that should work
Oct 26 2010 10:30PM
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BootySpunk.
Posts: 8148 - Karma: 84
Yah he did some weird shit but I haven't done trig stuff in forever except fucking trig subs in Calc and a bunch of integrals/derivatives for trig functions which is honestly the worst math shit ever. Besides sequences.
Oct 26 2010 10:32PM
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Bodhisattva
Posts: 878 - Karma: 138
thank you this helped me out a lot. i get it now that you explained it this way +k
Oct 26 2010 10:36PM
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Cthulhu
Posts: 528 - Karma: 21
i havent had to do math in 3 years and im boarderline retarded at it but trig was the easyest shit i have ever had to do, its just reapeating formulas
Oct 26 2010 10:37PM
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Gingeski
Posts: 100 - Karma: 16
He got one by multiplying the right hand side by 1+cosx, which is the reason the numerator on the right hand side becomes 1-cos^2x. (1+cosx)(1-cosx) "foils" to 1-cos^2x, or is just recognized as the factored form of a difference of two perfect squares.
Oct 26 2010 10:37PM
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Bodhisattva
Posts: 878 - Karma: 138
No need to do the work, I got it now. But I dont get it, my teacher said you can only work at one side at a time, i dont get how theyre using both sides
Oct 26 2010 10:39PM
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2_fadez
Posts: 7780 - Karma: 4,183
for the last step, when you get to: 1/1+cosx = 1-cosx/sin^2 multiply both sides by 1+cosx

now you've got 1 = (1-cosx)(1+cosx)/sin^2

multiply both of those together and you get: 1 = 1-cos^2/sin^2

now use the equation sin^2 + cos^2 = 1 to justify that 1-cos^2 = sin^2

so now you have 1 = sin^2/sin^2

thus 1=1
Oct 26 2010 10:39PM
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BootySpunk.
Posts: 8148 - Karma: 84
You didn't do it right...you were supposed to show that the left right equaled csc^2 not that 1 = 1 but I see how it makes sense what you did but I don't think it is what would be an acceptable answer even though it is technically right.
Oct 26 2010 10:41PM
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Gingeski
Posts: 100 - Karma: 16
yeah the teacher I had let us work on both sides, but there is one at my school that restricts it to only working on one side, figured your's was the same way
Oct 26 2010 10:41PM
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Mr.Gaga
Posts: 2021 - Karma: 542

There are many ways to do it, sometimes its easier only to work on one side of the equation, sometimes on the other, sometimes on both at the same time. Here is a exemple only working on the left side of the equation:

With csc^2=1+cot^2 and sin^2+cos^2=1:

1/(1+cos) + (csc^2 -1)/cos =         **(Note: csc^2=1+cot^2)

1/(1+cos) + cot^2/cos =

1/(1+cos) + cos/sin^2 =

sin^2/(1+cos)*(sin^2) + (1+cos)*cos/(1+cos)*(sin^2) =

sin^2 +cos^2 +cos / sin^2*(1+cos) =

(1+cos) /sin^2*(1+cos) =               **(Note: sin^2+cos^2=1)

1/sin^2= csc^2

Hope it will help. Good luck.

Oct 26 2010 10:52PM
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ARH
Posts: 117 - Karma: 14
the answer is you can't. 
Oct 27 2010 1:03AM
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