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?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Tangent line approximations with trig. functions

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