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Trigonometry Help

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    For small angles, the numerical value of sin theta is equivalent to tan theta. find the largest angle for which sine and tangent agree.

    2. Relevant equations
    I do not think a equation is need to answer this question.

    3. The attempt at a solution
    I honestly couldn't think of any way to solve this problem. My only attempt was to graph each of these general graphs, sin (theta) and tan (theta) and see where they intersect at.
  2. jcsd
  3. Sep 6, 2009 #2
    So you need to solve the following equation: sin x = tan x. Hint: rewrite tan x in terms of sin x and cos x and then see what you can do.
  4. Sep 6, 2009 #3
    Okay Thanks. So I computed sin x = sin x / cos x and cross multiplied and got sinxcosx= sinx. I don't know if I can keep on solving or use an trig identity from here.
  5. Sep 6, 2009 #4
    You can continue to solve algebraically. There are three cases:

    1. sin x = 0
    2. cos x = 1 (you are dividing by sin x assuming you're not in case 1)
    3. cos x = sin x = 0

    We can throw out case 3, so you only need to consider cases 1 and 2. Now you must find the largest angle that satisfies one of these equations. I'm guessing your teacher is looking for an angle between 0 and 2pi or else there wouldn't be an answer because there is no "largest angle."
  6. Sep 6, 2009 #5
    Yeah that's correct it has to be between 0 and 2 pi. But I don't understand how these cases can help in determining an angle. Since the problem is asking for when sin = tan and not sin = cos
  7. Sep 6, 2009 #6
    You are not finding where sin x = cos x. By using algebra, you have found that the set of solutions to the equation sin x = tan x is the union of the set of solutions to the equation sin x = 0 and the set of solutions to the equation cos x = 1. Now you simply want to find the largest angle in the resulting set.

    Set of solutions to sin x = 0 : {0, pi, 2pi}
    Set of solutions to cos x = 1 : {0, 2pi}
    Set of solutions to sin x = tan x : {0, pi, 2pi}

    Finding the answer should be simple now. Does this all make sense? You can think about it geometrically if that's easier. Try drawing some angles out and see if this answer makes sense!
  8. Sep 6, 2009 #7


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    Hi RaPiD! :smile:

    (have a theta: θ :wink:)
    Hi RaPiD! :smile:

    The magnitude of tan is always greater than that of sin (unless cos = 1), since tan = sin/cos, and cos < 1.

    So they can only be approximately equal.

    When the question says "for which sine and tangent agree", how much agreement (eg, how many decimal places) does it mean?
  9. Sep 7, 2009 #8
    Hi tiny-tim

    The question states to two decimal places.
  10. Sep 7, 2009 #9


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    ok, then the question is what is the maximum θ for which tanθ and sinθ are the same to two decimal places … just plug numbers into your calculator until you find it!

    (the alternative would be to solve tanθ - sinθ = .05, but that wouldn't exactly answer the question)
  11. Sep 7, 2009 #10
    hmm..is there a more mathematical way in solving this besides just plugging in. How wouldn't solving tanx - sinx = .05 wouldn't exactly the question. How did you even get .05?
  12. Sep 7, 2009 #11


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    oops, I meant .005 :redface:

    well, for example if tan = .176 and sin = .173, then to two decimal places they're .18 and .17, which is different.
  13. Sep 7, 2009 #12
    Yeah. I understand the concept of the question by which tan and sin are equal. It just that's going to be a lot of guessing and checking to fulfill the requirements of it being a small angle to two decimal places which I just realized has to be radians. :).
  14. Sep 7, 2009 #13
    Would solving sinx = tanx be a better approach for solving it?
  15. Sep 7, 2009 #14


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    No!!! sinx = tanx only for x = 0 (or a multiple of π).
  16. Sep 7, 2009 #15


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    You might try comparing the polynomial expansion of the 2 functions.
  17. Sep 7, 2009 #16
    Oh. So basically I still have to just plug in numbers into the calculator. There's no expression I could solve
  18. Sep 7, 2009 #17
    I would I go about doing a polynomial expansion?
  19. Sep 7, 2009 #18


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    google is a good place to start.

    Remember that you are looking for DIFFERENCE between sin and tan so trying to find where they are equal will not yield your solution. If you want an algebraic solution then the polynomial expansion is a good way to go. However if you are not familiar the expansions (you can look them up, or) you can plot the difference of the functions. From the graph you can learn when it is less then some given value.
    Last edited: Sep 7, 2009
  20. Sep 7, 2009 #19
    Okay thanks Integral. I will look up expansions on google and see where that takes me.
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