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Finding reciprocals

  1. Sep 27, 2008 #1

    csc theta=


    ctn theta=

    find theta
  2. jcsd
  3. Sep 27, 2008 #2


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    You need to show us some work.

    Here are some questions to get you moving.

    What is sin [itex] \theta [/itex] ?

    What is Tan [itex] \theta [/itex] ?

    Now use what you know about the definition of sin and Tan to draw a triangle. Apply Pythagorean theorem.
  4. Sep 27, 2008 #3


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    First you need to solve for x. That means making use of a trigo identity to get theta out of the way. Express sin theta and cos theta in terms of x, then apply that trigo identity.
  5. Sep 28, 2008 #4


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    Trig identities are one way to shift from one function to another (for example, if you know sec(x)= a/b and want to find tan(x), write a trig identity that converts sec to tan, say tan2(x)= sec2(x)- 1 and so that tan2(x)= x2- 1 and tan(x)= \sqrt{a^2/b^2- 1}= .\sqrt{a^2- b^2}/b [/itex].

    Still another, perhaps simpler, method is to draw a triangle with one angle x and write in appropriate lengths for the sides. In the above example, if sec(x)= a/b then, since sec= 1/cos or "hypotenuse over near side", our triangle would have hypotenuse a and near side b. By the Pythagorean theorem, the "opposite side" has length [itex]\sqrt{a^2- b^2}[/itex] and so tan(x), "opposite side over near side", is [itex]\sqrt{a^2- b^2}/b[/itex]

    But don't use the word "reciprocals" here. Reciprocals are specifically the "multiplicative inverses". Here you are concerned with "functional inverses".
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