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Simple square root factoring question

  1. Oct 4, 2004 #1

    is [tex]\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}?[/tex]

    If not, what should it be? :confused:

    Appreciate the help! :smile:
    Last edited: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2
    That is correct.
  4. Oct 4, 2004 #3
    Thank you for the help Sirus!
  5. Oct 4, 2004 #4


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    It is correct within a sign!
  6. Oct 4, 2004 #5
    What do you mean?
  7. Oct 4, 2004 #6
    No, that is not quite correct if you think of how a secondary definition of the modulus (absolute value) follows from the square root of a square,

    [tex]\sqrt{x^2} = \|x\|[/tex]

    In your case

    [tex]\sqrt{a^2-a^2\sin^2{x}} = a\cos{x}[/tex]

    can be written if and only if a and cos(x) are both positive or both negative; before you brought them out of the square root sign, you had the intermediate step,

    [tex]\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}}[/tex]


    [tex]\sqrt{a^2-a^2\sin^2{x}} = \sqrt{a^2\cos^2{x}} = \|a\cos x\|[/tex]

    which should hold true anyway since the left hand side is the positive square root.

    In general however, you can write it as a cos x if you have no major problems with the signs (you won't have any if theta lies between 0 and pi/2 and a > 0 for instance). But if you're proving something which involves this substitution, you had rather take it into account.

  8. Oct 5, 2004 #7
    I stand corrected. I forgot about that.
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