1. ## Linear Approximation question

I'm currently having a hard time with this question:

Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

I'm familiar with the linear approximation formula, but this is confusing me. I tried getting (with 29.1 as a "nice" point): (cos29.1)(x-29)(sin29.1) but that is incorrect. I think it has something to do with using this number and plugging it into the equation for a tangent line? I'm really not sure. Thanks in advance for any help!

2. Originally Posted by fleabass123
I'm currently having a hard time with this question:

Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

I'm familiar with the linear approximation formula, but this is confusing me. I tried getting (with 29.1 as a "nice" point): (cos29.1)(x-29)(sin29.1) but that is incorrect. I think it has something to do with using this number and plugging it into the equation for a tangent line? I'm really not sure. Thanks in advance for any help!

$30^\circ$ is $\frac{\pi}{6}$ radians

$29^\circ$ is $\frac{29\pi}{180}$ radians

at $x = \frac{\pi}{6}$ , the tangent line to the curve $y = \sin{x}$ has a slope $y' = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

the tangent line equation is ...

$y - \frac{1}{2} = \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right)
$

using $x = \frac{29\pi}{180}$ for the approximation ...

$\sin(29^\circ) \approx y = \frac{\sqrt{3}}{2}\left(\frac{29\pi}{180} - \frac{\pi}{6}\right) + \frac{1}{2}$

3. Ah, I had the right idea (kind of). Thanks for showing your process. I understand how to do it now.