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Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians. I want to know why its true and how it actually works. I would appreciate someone's help
One way to look at it is to take it as the linear approximation by the first derivative. We have ##\left. \dfrac{d}{dx}\right|_{x=0}\tan(x)=(1+\tan^2(x))|_{x=0}=1## which means that the tangent function is locally approximated by ##x \longmapsto (\tan(x))_0' \cdot x =1 \cdot x##.Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians. I want to know why its true and how it actually works. I would appreciate someone's help
But why is it true for radians only? I solved on my calculator and this is what i saw :Did you draw a sketch? A small part of the circle can be approximated by a straight line.
It is similar to the approximations ##\cos(x) \approx 1## and ##\sin(x) \approx x## for small x, and you can derive the approximation of the tangent that way as well.
##\tan 0.12 \approx \tan 7° \approx [x + O(x^3)]_{at \, 0} = 0.12 \pm 0.002## which is close to ##0.12##.But why is it true for radians only? I solved on my calculator and this is what i saw :
Tan (0.12) = 0.12 (where 0.12 is radians)
But when x is in degrees
Tan(0.12) = 0.00209
What is happening here?please explain.
The series expansion of ##\tan x## is ##\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5+\cdots##, where the powers of ##x## get bigger and bigger. This means for any value of ##x > 1##, the higher power terms in the series will contribute proportionally more than the lower terms. So the approximation only really works well when ##x\ll 1##, because then the higher order terms die out quickly. If you use degrees instead of radians, then you're effectively using ##d = \frac{180}{\pi}x## and calculating ##\tan d## instead of ##\tan x##. Since ##\frac{180}{\pi}\approx 57.2##, you should expect to get ##\tan d \approx \frac{1}{57.2}d##. Doing a quick calculation:But why is it true for radians only?
No reputable book would say that tan(x) = x. The proper relationship is that ##\tan(x) \approx x## for values of x near 0 (and in radians).Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians.
Radians are the natural units for angles.But why is it true for radians only?
That is a much weaker statement than the ratio. It would also apply for ##2x=x+\alpha(x)##, for example, but approximating 2x as x is usually a bad idea.This immediately implies ##\tan x = x + \alpha (x) ##, where ##\alpha (x)\to 0 ## as ##x\to 0 ##. In other words, the closer you get to ##0 ##, the smaller the difference between ##\tan x ## and ##x ## becomes.
That argument does not work in general, see my example with 2x instead of tan(x).If ##\tan x = x + \alpha (x)##, then ##\frac{\tan x}{x} = 1 + o(x) ##.
That direction is fine but that is a much weaker statement on the right side if you just require ##\hat \alpha(x)## to go to 0.Conversely, if ##\lim_{x\to 0} \frac{\tan x}{x} = 1 ##, then ##\tan x = x(1+\alpha (x)) =: x + \hat{\alpha} (x) ##. We just label things differently, it seems.
Your mistake goes deeper than sloppiness, because your concept of a (linear) approximation by the first derivative missed the point. It is essential that the normed direction tends towards zero not just the remainder term. We have ##\tan(0+v) = \tan(0) + \tan'(0)\cdot v + r(v)## and ##\lim_{v \to 0}\dfrac{r(v)}{||v||} =0##. This is the reason why @mfb's counterexample works, if this is not the case. You cannot skip the denominator.You are correct. I should have made my remainder term more explicit, hence my remark why I like yours more.
This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)Radians are the natural units for angles.
If you take a circle with radius 1, then the circumference is ##2\pi##. And there are also ##2\pi## radians in a revolution. So, the arc length of a circle radius 1 is equal to the angle in radians. The tangent function is defined as the ratio of lengths of a right triangle ##\frac{opposite}{adjacent}##. If you take a right triangle with a small angle, the length of the opposite leg is very close to the arc length of the circle next to it, and the length of the adjacent leg is very close to the radius of the circle. So, the tangent is very close to the angle in radians.
The photo you posted is the reason that we discourage images of work.This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)
You left out a PB in the length of the arc. The arc length should be AB = x PB.This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)
Thanks for thatYou left out a PB in the length of the arc. The arc length should be AB = x PB.
In my example, I used a radius of 1, so PB = 1. But you don't have to make that assumption.