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Verify trig identity

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Use a graphing calculator to test whether the following is an identity. If it is an identity, verify it. If it is not an identity, find a value of x for which both sides are defined but not equal.

    [tex]\frac{cos(-x)}{sin(x)cot(-x)}[/tex]=1


    2. Relevant equations
    None


    3. The attempt at a solution
    Ok, plug in the left side for y1, right side for y2, obviously not an identity. The second part where it ask for a x value is where Im having trouble. I thought maybe simplify the left hand side and find a value for whatever that is that equals 1..

    Cos(-x)/sin(X)(1/-tan(X))
    Cos(x)/sin(x)(-cosx/sinx)
    Cos(x)/-cos(x)
    Cos(x)/1-sin(x)
    Cos(x)-Sin(x)cos(x)
    Factor out to get
    1-sin(x)
    ok so now I have 1-sin(x)=1
    -sin(x)=0
    so find the value where -sin(x)=0??

    The teachers answer is - Not an identity, x=[tex]\pi/4[/tex]

    Im clueless about the second part I guess..
     
  2. jcsd
  3. Dec 8, 2009 #2
    so, you simplified it to the form of Cos(x)/-cos(x). This just equals -1. (your next step was incorrect, though... how did you get from -cos(x) to 1-sin(x)?)
    So it obviously is not equal to one. The problem statement asks for "a value of x for which both sides are defined but not equal". We have just shown that whenever the expression is defined, the equation doesn't hold. So you just need to find a value where the expression is defined.
     
  4. Dec 8, 2009 #3
    Yeah your right should be -1, I'm not quite sure how I would go about finding a value where the expression is defined.

    Anyone? That was my original question, Im not sure how to find that value.
     
    Last edited: Dec 8, 2009
  5. Dec 9, 2009 #4
    It's defined almost everywhere. Since cot(-x)=-cos(x)/sin(x), you need to make sure sin(x)=/=0.

    Then you also need to make sure that the simplified expression, cos(x)/-cos(x) is defined, i.e. cos(x)=/=0.

    That's all. There are plenty of angles for which this holds. pi/4 is one, but you can also have pi/3, 2*pi/3, etc..
     
  6. Dec 9, 2009 #5

    Mentallic

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    Homework Helper

    Stevo6754, since you're able to simplify the expression and make it [tex]\frac{cos(-x)}{sinxcot(-x)}=-1[/tex], this is telling you that for all defined values of x (they are undefined where the denominator=0) it is not equal to 1. So like grief has said, the values you're looking for are not just [itex]\pi/4[/itex] but any other value that is defined.

    Oh and while [itex]cos^2x=1-sin^2x[/itex] this does not mean [itex]cosx=1-sinx[/itex] since, if you square both sides you'll get [itex]cos^2x=(1-sinx)^2=1-2sinx+sin^2x\neq 1-sin^2x[/itex]
     
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