Need help with trig identities problem

Discussion in 'Homework Help' started by strantor, Feb 27, 2013.

  1. strantor

    Thread Starter AAC Fanatic!

    Oct 3, 2010
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    PLZ NEED HLP URGENT.

    I kid, I kid. I had this question in my homework last week and tried probably 5 times and then wrote it off as a loss. Now it's come back around to abuse me on a take home quiz. So I've just now tried it another 7 or 8 times and about to pull my hair out. I'm supposed to pick either the left side or the right side, and using trig identities, work through the problem until the 2 sides are the same. I've tried both side and gotten nowhere.

    [​IMG]
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Your first line is ok.
    Don't expand the denominator.
    Expand the nominator, as you have done on the second line.
    Your result of 2 + 2sin is correct.
    The identity should be obvious from there.
     
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  3. MrChips

    Moderator

    Oct 2, 2009
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    Note that this identity is not true under two conditions.
     
  4. amilton542

    Active Member

    Nov 13, 2010
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     \frac{cos(x)}{1 + sin(x)} + \frac{1 + sin(x)}{cos{x}} =

     \frac{cos^2(x) + [1 + sin(x)]^2}{[cos(x)][1 + sin(x)]} =

     \frac{[cos^2(x) + sin^2(x)] + [1 + 2sin(x)]}{[cos(x)][1 + sin(x)]} =

     \frac{2 + 2sin(x)}{[cos(x)][1 + sin(x)]} =

     \frac{(2)(1 + sin(x))}{[cos(x)][1 + sin(x)]} = 2sec(x)
     
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  5. strantor

    Thread Starter AAC Fanatic!

    Oct 3, 2010
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    DOH! So simple. Thank you MrChips!

    Edit: and amilton too!
     
  6. amilton542

    Active Member

    Nov 13, 2010
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