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I Approximation of tan()

  1. Mar 31, 2017 #1
    I'm following this video:

    The professor says that for small angles, tan(Θ) = dy/dx. I don't understand why this is so. Tan(Θ) is equal to sin(Θ) / cos(Θ), and if Θ is small, then cos(Θ) is about 1, which means dx = 1, not a infinitesimally small number.
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  3. Mar 31, 2017 #2


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    Staff: Mentor

    That's not correct. Since ##\sin \theta \sim \theta##, then ##\tan \theta \sim \theta##, so you have that ##dy/dx \sim \theta##. It doesn't tell you anything about ##dx## by itself.

    You have to go back to the definition of ##tan \theta## for a right triangle.
  4. Mar 31, 2017 #3


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    dy and dx refer to changes, here in the curve. To express them with cosine and sine, you have to introduce the size of the triangle, and then let that size go to zero.
    The ratio of change of y divided by the ratio of change of x is dy/dx, and that is indeed the tangent of θ.
  5. Mar 31, 2017 #4

    The video starts with analyzing the forces in the Y direction, and comes out with -Tsin(theta) and + Tsin(theta + delta theta). So the sin, which is then approximated to tan, should give a ratio of forces. But the video says tan = dy/dx, meaning tan went from a ratio of forces to a ratio of distance. What?
  6. Mar 31, 2017 #5


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    What's wrong with that? A ratio is a ratio. A ratio of forces is a quantity without units. A ratio of distances is also a quantity without units. Is that what's bothering you?

    The lecturer is playing fast and loose with his notation. He wrote ##\sin(\theta) \approx \frac{\sin(\theta)}{\cos(\theta)}##, which is fine, since he stipulated that both ##\theta## and ##\Delta \theta## were small. He then wrote ##\frac{\sin(\theta)}{\cos(\theta)} \approx \tan(\theta)##, saying these were approximately equal. In fact, these two quantities are exactly equal by definition of the tangent function.
  7. Apr 1, 2017 #6
    Yes, that's it. Originally, theta was defined as the angle made by the force, and so I expected the sin(theta) function to be analyzed in terms of forces. As in, sin(theta) equals Y Tension divided by X Tension. But I was thrown off by how he made the correlation between DISTANCES. So I reconciled this by thinking that the tension is at the same angle as the string, so tan(theta) is equal to Y Tension / X tension as well as dy/dx.

    So if dy / dx is also equal to Y Tension / X Tension, is it valid to solve for the Wave Equation in this way? So instead of depending on the position of the string, it would be in terms of forces on the string?
  8. Apr 1, 2017 #7


    Staff: Mentor

    The work shown in the video deals exclusively with the forces on the string, not the position of either end. The only role that position plays is that the angles between the force and the horizontal is slightly different at each end.
  9. Apr 1, 2017 #8
    What I mean is that if sin is substituted with dy/dx, eventually the Wave Equation comes out with d2y/dx2 = ma. What if instead, we said sin approximately equals tan, and tan = TensionY1/TensionX1 for the left end, and the right end is TensionY2/TensionX2. Then (TensionY2/TensionX2 - TensionY1/TensionX1)(deltaX/deltaX) = d/dx of (TensionY/TensionX)
  10. Apr 2, 2017 #9


    Staff: Mentor

    What do you mean "what if"? That's exactly what the lecturer said.
    For one thing the angles are different at the two ends. Also, you need to clarify what your terms mean. In the video, the lecturer had T for the tension, which was the same at both ends. What do you mean by TensionX1 etc.?
  11. Apr 2, 2017 #10
    So sin(theta) is approximately tan(theta), and tan(theta) is Tension in Y direction / Tension in X1 direction. That's for the left side of the string.
    Same with the right side, but the tensions in each directions will be different as the angle is now theta+delta theta
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