Question

find the linerization of a suitable function and then use it to approximate the number:

sin 0.1

Answer 1

use sin(x) at 0. the equation is
\[y=cos(0)x+0\] or just
\[y=x\]

Answer 2

plug in .1 get .1

Answer 3

I am not sure I get it, but I guess you need the taylor series of sinx at 0. And if it is linear than you only need the T1.
the derivative of sin is cosx.
so T1=sin0+cos0 (x-0)=x

Answer 4

is that clear?

Answer 5

kinda

Answer 6

oh yes its clear all i wanted to know is the number which is a function of sine which is closest to 0.1 thanks andras and satellite!

Answer 7

this is the same as the question "find the equation for the line tangent to the curve of
\[y=\sin(x) \] at (0,0)

Answer 8

and undoubably you have been shown that
\[lim_{x->0}\frac{\sin(x)}{x}=1\] yes? this says that near 0,
\[\sin(x)\] looks like x. so
\[\sin(.1)=.1\] approximately

Answer 9

like this approach satellite!

Answer 10

like how sin 0 = 0?

Answer 11

like how sin 0 = 0?

Answer 12

well it is true that
\[\sin(0)=0\] but more to the point
\[lim_{x->0}\frac{\sin(x)}{x}=1\] they say different things. the first just evaluates the function. the second says that near 0 the ratio of sin(x) to x is 1. this means sin(x) is close to x if x is close to 0

Answer 13

okay thank you satellite