# 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 $$\frac{\pi}{6}$$)