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Finding the Equation of a Tangent

  1. May 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent at the point indicated
    y=3cscx
    x=pi/4

    2. The attempt at a solution

    So to do the question you need a point (which im given), the slope and then you need to substitute that all into y=mx+b.

    I believe I have differentiated correctly with

    y=3(-cot(x)csc(x))

    Then I find the slope by inserting pi/4 into the differentiated equation

    y=3(-cot(pi/4)csc(pi/4))

    After this is done I put everything into y=mx+b, and this is where my problem is. Is there anyways to simplify 3csc(pi/4) and 3(-cot(pi/4)csc(pi/4))? I think I remember doing something similar in math a long time ago but I forgot now. The final answer does not include -cot or csc so there must be a way I dont remember.
     
  2. jcsd
  3. May 30, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    What is cosec and cotangent in terms of the other trig functions?

    1/(sin x) = csc x
    1/( tan x) = cot x

    So 3 csc (pi/4) = [itex]\frac{3}{\sin (\pi/4)}[/itex].

    The exact value for sin (pi/4) is 1/(sqrt2) so..
    [itex]3 \csc (\pi/4) = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}[/itex]

    So the same for cot, knowing tan (pi/4) = 1.
     
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