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Tangent line approximations with trig. functions

  1. Oct 19, 2006 #1
    Find the equations of the tangent lines to the graph of [tex] f(x) = \sin x [/tex] at [tex] x = 0[/tex] and at [tex] x = \frac{ \pi }{3}[/tex]. Use each tangent line to approximate [tex] \sin( \frac{ \pi }{6} ) [/tex]. Would you expect these results to be equally accurate, since they are taken equally far away from [tex] x = \frac{ \pi }{6} [/tex] but on opposite sides? If the accuracy is different, can you account for the difference?

    Here are my attempts:
    Well, first let's find the equations of the tangent lines.
    [tex] f(0) = \sin (0) = 0[/tex]
    [tex] f(\frac{ \pi }{3}) = \sin( \frac{ \pi }{3}) = \frac{\sqrt{3}}{2} [/tex]

    [tex] f'(x) = \cos (x) [/tex]
    [tex] f'(0) = \cos (0) = 1 [/tex]
    [tex] f'(\frac{ \pi }{3}) = \cos( \frac{ \pi }{3}) = \frac{1}{2} [/tex]

    So we have (0,0), m = 1.
    y = x <---- tangent line at x = 0.

    And we have [tex] (\frac{ \pi }{3} , \frac{ \sqrt{3}}{2}) , m = \frac{1}{2} [/tex]
    [tex]y = \frac{1}{2} x + \frac{3\sqrt{3} - \pi}{6} [/tex] <---- tangent line at [tex] x = \frac{ \pi }{3} [/tex]

    Now we'll use the tangent lines to approximate [tex] \sin( \frac{ \pi }{6})[/tex].
    Using y = x.
    [tex]y(\frac{ \pi }{6}) = \frac{ \pi }{6} \approx 0.5236[/tex]

    Using [tex] y = \frac{1}{2} x + \frac{3\sqrt{3} - \pi}{6} [/tex]
    [tex]y(\frac{ \pi }{6}) = \frac{ \sqrt{3} }{2} - \frac{ \pi }{12} \approx 0.6042 [/tex]

    They are obviously different. I don't know why they are different though.

    Thanks in advance.
    Last edited: Oct 19, 2006
  2. jcsd
  3. Oct 19, 2006 #2
    [tex] L(x) = f(a) +f'(a)(x-a) [/tex]. Think about the graph of the sine function. The linear approximation of [tex] \sin x [/tex] at [tex] a = 0 [/tex] is [tex] \sin x \approx x [/tex]. Also the tangent line approximations are taken from opposite sides of [tex] \sin \frac{\pi}{6} [/tex]. So, its similar to the squeeze theorem. One approximation will overestimate the true value, and another approximation will underestimate the true value.
    Last edited: Oct 19, 2006
  4. Oct 19, 2006 #3
    well, the thing that bothers me, is you said that one approximation should be an underestimate and one should be an overestimate.

    the true value of [tex] \sin( \frac{ \pi }{6}) = \frac{1}{2} [/tex]

    1/2 is not between our two approximations.
  5. Oct 19, 2006 #4
    ah yes my mistake. I think the key reason is the difference between the areas of approximation (i.e. 0 and [tex] \frac{\pi}{6} [/tex])
    Last edited: Oct 19, 2006
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