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How to find algebraic expressions?

  1. Feb 20, 2007 #1
    How do you find an equivalent expression free of trig funtions for g(b)=cos(arctan(b))? :confused:
  2. jcsd
  3. Feb 20, 2007 #2


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    Draw a right angled triangle, with one of the non-hypoteneuse sides having length b, and the other non-hypoteneuse having length 1.
  4. Feb 21, 2007 #3


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    A little more detail on AKG's suggestion: You are looking for [itex]cos(\theta)[/itex] with [itex]\theta= arctan(b)[/itex] or [itex]tan(\theta)= b[/itex].

    Since "[itex]tan(\theta)[/itex]" is "opposite side over near side" you can think of b as b/1. Draw a right triangle have "opposite side" b and "near side" 1. Now [itex]cos(\theta)[/itex] is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenuse and find [itex]cos(\theta)[/itex].
  5. Feb 21, 2007 #4


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    Alternatively, remember the identity: [itex]\sec^{2}(\theta)=\tan^{2}(\theta)+1[/itex]
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