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anonymous

  • 5 years ago

i need to solve these two problem using identities...sin2x+cos2x=0 and csc^2x(x/2)=2secx . help on either or both would be much appreciated!

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  1. anonymous
    • 5 years ago
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    michelle i'm not sure if these identifies even exist.... i plugged in values for them and they don't work out. sin(2pi) + cos(2pi) = 1 though according to the ID you provided, it should be zero and the second..i'm not sure what it says is it csc(x/2)^(2x) = 2 secx?

  2. anonymous
    • 5 years ago
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    we're supposed to be solving for x using identities. sorry the second is confusing...its supposed to be\[ \csc ^{2}(x/2)=2secx\]

  3. anonymous
    • 5 years ago
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    that should be an equal sign...it doesnt look like it on my computer, but it might just be screwy

  4. anonymous
    • 5 years ago
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    okay michelle, i kinda gave up on the first and went straight to the second i really don't think that these exist because i've done it several ways, and I keep getting the same answer if i got different answers, i might be doing it wrong. but i get the same answer i'll do it out here and you can decide for yourself \[\csc^2 ({\frac{x}{2}}) = \frac{2}{\cos x}\] \[\frac{1}{\sin ^2 (\frac{x}{2})} = \frac{2}{\cos x}\] \[2\sin^2(\frac{x}{2}) = \cos x\] \[2(\frac{1 - \cos x}{2}) = \cos x\] \[1 - \cos x = \cos x\] \[1 = 2 \cos x\] \[\cos x = \frac{1}{2}\] ^-- i keep getting that T-T

  5. anonymous
    • 5 years ago
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    thats what we're supposed to get! thanks so much...that helps a lot. we're solving for x. and i can solve for cosx=1/2, because we know that x=pi/3 or 5pi/3. thats what we're supposed to be doing =)

  6. anonymous
    • 5 years ago
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    AHHHH MICHELLE!!! THAT MEANS THESE AREN'T IDENTITIES!!! RAWRRRRRRRRRRRRRRR you had me tearing my hair out trying to get the equation to equal.. T-T sighs give me a bit more time, i'll do the first one

  7. anonymous
    • 5 years ago
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    i'll try doing the first one*** i'm not sure if i'll be able to

  8. anonymous
    • 5 years ago
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    omg im so sorry. i meant we're solving for x like, using the different identities. idk...thats just what it says on the top of my sheet..."solve for x usiing identities". im sorry...you really dont have to. thanks for that one tho =)

  9. anonymous
    • 5 years ago
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    lol so whos up to what here ?

  10. anonymous
    • 5 years ago
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    i got the second one see if you can do the first one solve for x

  11. anonymous
    • 5 years ago
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    yeh , its relatively easy lol

  12. anonymous
    • 5 years ago
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    what i've got so far, i think is right \[2 \sin x * \cos x + \cos^2 x - \sin^2 x = 0\] \[2 \tan + 1 - \tan^2 x = 0\] \[\tan^2 x - 2 \tan x - 1 = 0\] now you have to solve for x using the quadratic formula, which i dont know how to take into account the tangent

  13. anonymous
    • 5 years ago
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    its a perfect square

  14. anonymous
    • 5 years ago
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    you made a mistake between the 2nd last and the last line a +1 becoame a -1

  15. anonymous
    • 5 years ago
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    no, i negated the entire thing, it is NOT a perfect square

  16. anonymous
    • 5 years ago
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    you are doing it the hard way anyway

  17. anonymous
    • 5 years ago
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    wait im confused. how did you get from the first step to the second step (with the tans)? are there identities that tell you that? i might just not know them

  18. anonymous
    • 5 years ago
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    then, please, tell us how to do it...

  19. anonymous
    • 5 years ago
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    i divided the entire equation by sin^2(x)

  20. anonymous
    • 5 years ago
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    ohhh gotcha.

  21. anonymous
    • 5 years ago
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    sin(2x) = -cos(2x) tan(2x) =-1 related angle of pi/4 tan is negative in the 2nd and 4th quadrants now we originally had 0<=x<=2pi , multiply through by 2 ( because we have (2x) as the argument of our tan )

  22. anonymous
    • 5 years ago
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    *facepalm*

  23. anonymous
    • 5 years ago
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    so 0<=x<=4pi ( ie we need two revolutions

  24. anonymous
    • 5 years ago
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    trig id overboard.... michelle, listen to elec, he's right

  25. anonymous
    • 5 years ago
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    so 2x = 3pi/4 , 7pi/4 ( from the first revolution )

  26. anonymous
    • 5 years ago
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    now, we need another revolution, so add 2pi to each of these

  27. anonymous
    • 5 years ago
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    so 2x = 3pi/4 , 7pi/4 , 11pi/ 4 , 15pi/ 4 then divide by 2

  28. anonymous
    • 5 years ago
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    x = \[x = (3\pi/8 ) , (7\pi/8) , (11\pi/8), (15\pi/8) \]

  29. anonymous
    • 5 years ago
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    wow. thank you so so so much. you guys are both like, geniuses. thanks =)) i gave you both medals, whatever that means.

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