Homework Help: Tangent line approximations with trig. functions

1. Oct 19, 2006

Jacobpm64

Find the equations of the tangent lines to the graph of $$f(x) = \sin x$$ at $$x = 0$$ and at $$x = \frac{ \pi }{3}$$. Use each tangent line to approximate $$\sin( \frac{ \pi }{6} )$$. Would you expect these results to be equally accurate, since they are taken equally far away from $$x = \frac{ \pi }{6}$$ 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.
$$f(0) = \sin (0) = 0$$
$$f(\frac{ \pi }{3}) = \sin( \frac{ \pi }{3}) = \frac{\sqrt{3}}{2}$$

$$f'(x) = \cos (x)$$
$$f'(0) = \cos (0) = 1$$
$$f'(\frac{ \pi }{3}) = \cos( \frac{ \pi }{3}) = \frac{1}{2}$$

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

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

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

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

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

Last edited: Oct 19, 2006
2. Oct 19, 2006

$$L(x) = f(a) +f'(a)(x-a)$$. Think about the graph of the sine function. The linear approximation of $$\sin x$$ at $$a = 0$$ is $$\sin x \approx x$$. Also the tangent line approximations are taken from opposite sides of $$\sin \frac{\pi}{6}$$. 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
3. Oct 19, 2006

Jacobpm64

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 $$\sin( \frac{ \pi }{6}) = \frac{1}{2}$$

1/2 is not between our two approximations.

4. Oct 19, 2006

ah yes my mistake. I think the key reason is the difference between the areas of approximation (i.e. 0 and $$\frac{\pi}{6}$$)