Question

why when x approches 0 ( x very small angle) we consider almost cos(x)=1- (x^2)/2

Answer 1

Well the reason is because 1-(x^2)/2 and cos(x) are the same at x=0 and are also the same at x=0 for their first, second, and third derivatives.

Answer 2

In fact, if you were to make a polynomial with an infinite amount of terms that follows this sort of pattern it would become cos(x). \[\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\]

Answer 3

yeah but i think it comes from this limit \[\lim (\cos(\theta)-1)/x = 0\] but idk

Answer 4

It's related, but the real reason why you can use the approximation is because all that 1-x^2/2 is is simply a polynomial approximation to the cosine function that was specifically made to have the same value and first, second, and third derivative at x=0 which by extension will give similar results for other small values near 0. But you can very easily extend this like I described to have a curve that also has the same higher order derivatives as well. If two curves have the same derivatives and higher order derivatives at a point for all derivatives then they're the same.

Answer 5

Since the infinite polynomial is the same as cos(x) it becomes apparent why that limit is also true since they're just different representations of the same things. If you just subtract off 1 from it and then divide it by x, the lowest term is the x^2/2 term and it becomes x/2 which when you plug in 0 becomes simply 0 along with all the other infinite number of higher terms.

Answer 6

i think i found it we have this limit \[\lim_{x \rightarrow 0}\frac{ 1-\cos(x) }{ \frac{ x^2 }{ 2} }= 1/2\] and thenu move that x to the right u ll find it

Answer 7

yeah i think that way too will work thx kainui

Answer 8

that limit its not x^2/2 in the dominator its just x^2 ok