and i think the value is less than 1 right? certainly sin1 ≠ 1. here units i used are degrees.
In: 6
First thing, your units *must* be radians. It doesn’t work otherwise. With that in mind, it entirely depends on the accuracy you want. To get an actual answer, you would need to specify the accuracy.
For example, if you want to approximate sin(x) as x with 90% accuracy, you can use that approximation up to nearly 45° (so in radians, sin(pi/4) is about 90% of pi/4).
If you want 95% accuracy, it works up to about pi/6, or 30°. If you want 99% accuracy, it only works up to about 14°, about 0.245 radians.
>sinθ≈θ
Isn’t exactly right. It’s just what we tell younger students, so that we don’t have to teach them perturbation theory.
Really the statement should be
>sinθ~θ, for θ –> 0
Which is a shorthand for saying that sinθ/θ tends toward 1 as θ tends toward 0.
It isn’t really saying that these two things are “approximately equal”. Rather, the fact that the ratio between the two gets closer and closer to 1, indicates that they are “changing in a similar way”, so the isn’t much space for the two to differ.
EDIT:
In similar fashion your teachers will likely say “much less than” and use a double < symbol to represent it
> ” << ”
This is also a cheeky adaptation of notation from the same field of mathematics.
EDIT EDIT:
As others have said, sin needs to be in radians, but that is generally considered to be the default assumption when working with sin in any mathematical context.
EDIT EDIT EDIT:
If you are computationally minded person, you may know “big o notation”.
>sinθ~θ, for θ –> 0
is just more stringent version of:
>sinθ=O(θ), for θ –> 0
EDIT IV: Not sure why I’m getting down voted?
EDIT V: Crap I’m on explain like I’m five -_-
It depends on your accuracy. Firstly this rule applies to radiens and not for degrees. So sin 1 = 0.84 ~ 1, at least with 80% accuracy. With x=.1 you get an accuracy of 99.8%
Firstly, you need to be in radians. Sin x ≠ x if x is in degrees. Second, the approximation gets better for smaller angles, so the point where it becomes acceptable will depend on how much accuracy you need. Sin(1) = 0.84, which might be ok, depending what you need it for. Sin(0.5) = 0.47 is pretty good. Sin(0.1) = 0.0999 which is basically equal for most purposes
The Taylor series for sin is sinx = x – x^(3)/3! + x^(5)/5! – x^(7)/7! +…
This means you can approximate sinx by taking a few elements from the series – the more elements you take, the more accurate the value. If you take just one element (x) then the error will be no more than |x^(3)|/3! so for example if x=1 then the error is no more than 1/6. Of course the larger x is, the larger the error.
think of a smallest number you can think of. like super small. theta is about that small plus minus epsilon, another small number.
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As other answer had said, θ must be in radian. This is because θ actually represent a ratio of length of an circle arc to the radius; while sine represents the ratio of half the chord to the radius. If you draw them on a picture, you will notice that these 2 lengths should be similar to each other: you are comparing the length of a chord to the arc length of the circle extending it.
When you do precise mathematics, saying “approximately” is no longer sufficient, you need to demonstrate how small the error is. Except for the case θ=0, sin(θ) is never equal to θ and there is always a small error, so you need to estimate it. As it turns out, the error is cubic in θ. That is, the absolute difference between θ and sin(θ) is never more than θ^3 . For example, if θ<=0.1, you can guarantee that the error is no more than 0.001, so you guarantee 3 significant figures.
How you we estimate this? It is a theorem from calculus called Taylor’s remainder theorem. In this case, it says that if you use θ to approximate sin(θ), then the error equals -θ^3 cos(t) for some t between 0 and θ, and even though we don’t know what exactly is t, we can guarantee that |cos(t)|<=1. The above estimate come from Taylor’s 3rd order approximation.
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