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Trigonometry Help!

  1. Aug 18, 2005 #1
    I am having extreme difficulty with my Trigonometery homework.

    In particular, When I simplify the expression:
    I come up with cos2x.

    This is obviously incorrect. Where am I going wrong?
  2. jcsd
  3. Aug 18, 2005 #2


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    Well it depends on how you want to simplify. Do you need to get rid of the double angles or just to 'shortes' possible way of expression?

    [itex]\cos{2x}[/itex] is indeed wrong...

    You could factor [itex]\cos{2x}[/itex], don't know if that's what you want.
    You could also use the double angle formula backwards on [itex]\cos{2x}\sin{2x}[/itex] to get [itex]\frac{\sin{4x}}{2}[/itex]
  4. Aug 18, 2005 #3
    I should have written it:

    cos(squared) x sin(squared) x - cos(squared) x

    shortest way
  5. Aug 18, 2005 #4


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    What do you mean? Now you say [itex]\cos ^2 x\sin ^2 x - \cos ^2 x[/itex].
    That's not the same as your initial expressions :confused:

    You mean that's what you actually meant? What did you try already then?
  6. Aug 18, 2005 #5
    Still trying figure out how to get all of the expressions to translate into typed text....sorry.

    I cos(sq'd)x-cos(sq'd)x
    then 1-sin(sq'd)=cos(sq'd)X

    this is why I came up with cos(sq'd)x
  7. Aug 18, 2005 #6


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    Well, I'd do about the same but you get something different:

    [tex]\cos ^2 x\sin ^2 x - \cos ^2 x = \cos ^2 x\left( {1 - \cos ^2 x} \right) - \cos ^2 x = \cos ^2 x - \cos ^4 x - \cos ^2 x = - \cos ^4 x[/tex]
  8. Aug 18, 2005 #7
    I see the step I missed. Thank you very much.
    School is hard when you have been out of it for 18 years. :smile:
  9. Aug 18, 2005 #8


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    I can imagine you need some freshing up :wink:
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