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Trig Substitution

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    trig_zps06f1e0b1.png


    2. Relevant equations



    3. The attempt at a solution
    This isn't really a traditional question, but can someone explain to me how substituting u = tan^-1(x/y) got to that final value? I'm trying to understand this for an exam coming up.
     
  2. jcsd
  3. Oct 2, 2013 #2

    CompuChip

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    I didn't check, but using these identities
    $$\sin(\tan^{-1}(z)) = \frac{z}{\sqrt{z^2 + 1}}$$
    and
    $$\cos(\tan^{-1}(z)) = \frac{1}{\sqrt{z^2 + 1}}$$
    should get you started.
     
  4. Oct 2, 2013 #3

    arildno

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    It is a good exercise to derive the simplified expressions of trigs with inverse trigs as their arguments.
    For example, let's take cos(asin(x)). Not bothering about the precise domain for the inverse trig right now, we know that we have the identity:
    [tex]\cos^{2}(asin(x))+\sin^{2}(asin(x))=1[/tex]
    But, the latter term on the LHS simplifes to x^2!!
    Thus, we have:
    [tex]\cos(asin(x))=\pm\sqrt{1-x^{2}}[/tex]

    This is also readily seen geometrically:
    If we look at a right-angled triangle with unit hypotenuse, and sine equal to x (to which the relevant angle is asin(x)), then that expression falls right out of the Pythagorean theorem.
     
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