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Homework Help: Question about sin rules

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I have the equation (sin(2x))/x = ?

    3. The attempt at a solution

    I know that the answer to this is 2, but im not sure why (sin(2x))/x = 2

    Can somone explain?

  2. jcsd
  3. Oct 8, 2012 #2


    Staff: Mentor

    (sin(2x))/x ≠ 2, so perhaps you are leaving something out of the problem. What is the complete problem statement?
  4. Oct 8, 2012 #3
    Find the value of the constant a for which the function below is continuous everywhere. Fully
    explain your reasoning.

    ........... a+x2 while x≤0
    f(x) = {
    ........... (sin(2x))/x while x>0
  5. Oct 8, 2012 #4


    Staff: Mentor

    The problem is really asking about limits, namely
    $$ \lim_{x \to 0}\frac{sin(2x)}{x}$$

    Do you know any other limits that involve trig functions?
  6. Oct 8, 2012 #5
    I'm not sure what you mean by your question.

    I know that $$ \lim_{x \to 0}\frac{sin(x)}{x}$$ is equal to 1

    I also know that the limit as x approaches 0 from the negative and the limit at x=0 are both just equal to a.

    So this means that a just equals $$ \lim_{x \to 0}\frac{sin(2x)}{x}$$

    I know the answer to the question is 2 so that means $$ \lim_{x \to 0}\frac{sin(2x)}{x}$$ must equal 2 but im not sure how to do it.
  7. Oct 8, 2012 #6


    Staff: Mentor

    There are at least a couple of ways to go.
    1) Double angle identity for sine
    2) Adjust things so that you have sin(2x)/(2x) times some other stuff.
  8. Oct 8, 2012 #7
    For 2) do you mean like


    =so you get 1*2


    Would that work?
  9. Oct 8, 2012 #8
    (sin(2x))/ x = (sin2(1))/ 1
    = sin(2)(1)/1
    = sin(2)
    = 0.0349
  10. Oct 8, 2012 #9


    User Avatar
    Homework Helper

    That's the limit as x goes to 1, not 0.

    Yep, exactly! This would be the simplest way to do it, so if you ever get a question like

    [tex]\lim_{x\to 0}\frac{\sin(ax)}{b}[/tex] then this is equivalent to
    [tex]\lim_{x\to 0}\frac{a}{b}\cdot\frac{\sin(ax)}{a}=\frac{a}{b}[/tex]
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