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Trig identities help! MATHMETICIANS

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dave44
Posts: 11052 - Karma: 113
Okay, I feel like a retard. Did this shit like a year ago, but I need to recap and can't remember how to do this for the life of me (should be very simple).

Find the values of sin(x) and cos(x) when cos(2x) is 1/9
Oct 28 2010 10:04PM
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Squid.
Posts: 1904 - Karma: 23
ti 89

solve(cos^(-1)(1/9)=2x,x)

boom
Oct 28 2010 10:12PM
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Gingeski
Posts: 100 - Karma: 16
Non-calculator way:
Oct 28 2010 10:43PM
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Gingeski
Posts: 100 - Karma: 16
Oops, accidently hit enter. Cos2x=cos^2x-sin^2x Also cos2x=2cos^2x-1. And cos^2x=1-2sin^2x Set the second two both equal to 1/9, should get cosx=square root of 5/9 and sinx=2/3. Substitute answers back into first equation to check answers. 1/9= 5/9-4/9
Oct 28 2010 10:47PM
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skichicky3
Posts: 2593 - Karma: 64
SOHCAHTOA

Cos (2x) = 1/9 = adjacent/ hypotenuse

2x= 1/9, x= (1/9)/2 =1/18

x= 1/18

sin (1/18) = .0556
cos (1/18)= .9985

hope that helps, since I just basically did that whole thing for you. I'm a math nerd.


Oct 28 2010 10:49PM
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1337
Posts: 31038 - Karma: 29,003
answers are correct
Oct 28 2010 10:50PM
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skichicky3
Posts: 2593 - Karma: 64
Did you not see where I said I'm a math nerd! YOU should know better Jamie!
Oct 28 2010 10:52PM
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1337
Posts: 31038 - Karma: 29,003
I came here to write the same thing, but just backed you up. And that makes two of us kimber :)
Oct 28 2010 10:57PM
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Gingeski
Posts: 100 - Karma: 16
How is that correct? The cos of cos2x cannot just disappear, you'd have to take the cosine inverse to get rid of it. So to solve for x, it would be
cos2x=1/9
cos^-1 (cos2x)=cos^-1 (1/9)
2x=cos^-1 (1/9)
x=cos^-1 (1/18)

But the problem can easily be solved without a calculator using double angle identities, so why not just do it that way and get "nice" numbers?
Oct 28 2010 11:16PM
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Squid.
Posts: 1904 - Karma: 23
lol not even close
Oct 29 2010 9:42AM
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skichicky3
Posts: 2593 - Karma: 64
lol you are right, I completely forgot afterwards that cosx does not equal cos2x/2

I fail, maybe I should go back to high school
Oct 29 2010 11:09AM
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