[RESOLVED] Since an angle is too small

[eranga262154]

Hi all,

Assume that an angle α. Since it is too small (α <<0) we can say that Sin α = α, Tan α = α and Cos α = 1- ((α^2)/2).

How can we proof those three.

[VBAhack]

I believe the argument that's normally used involves looking at the series expansion. For example:

sin x = x - x³/3! + x⁵/5! - x⁷/7! + . . .

if x is sufficiently small all terms except the 1st are negligible. A simple calculation will bear this out. A similar story can be created for the other trig functions. At least this is how I remember it. I don't know if this can be considered a 'proof' though.

[krtxmrtz]

Hi all,

Assume that an angle α. Since it is too small (α <<0) we can say that Sin α = α, Tan α = α and Cos α = 1- ((α^2)/2).

How can we proof those three.

If you want a way to prove that sin(a) = a for small a, take a look at this:

http://www.vbforums.com/showpost.php...52&postcount=8

It's based on geometric considerations and doesn't require the Taylor series.

[eranga262154]

I believe the argument that's normally used involves looking at the series expansion. For example:

sin x = x - x³/3! + x⁵/5! - x⁷/7! + . . .

if x is sufficiently small all terms except the 1st are negligible. A simple calculation will bear this out. A similar story can be created for the other trig functions. At least this is how I remember it. I don't know if this can be considered a 'proof' though.

Ya, first I think it is the best way to do it. Since x is too small only first term (ie x) is the remaining part.

Anyway thanks for your advice.

[eranga262154]

It's based on geometric considerations and doesn't require the Taylor series.

Thank,
I go through your link. It is ok.

But I think it is easy to used Taylor series. Is it ok?

[krtxmrtz]

Thank,
I go through your link. It is ok.

But I think it is easy to used Taylor series. Is it ok?

Sure it's easier but the point is, the Taylor series can't be derived without the knowledge of derivatives and derivation. And in order to find the derivative of the sine you must calculate the limit lim (h -> 0) [sin(x+h) - sin(x)] / h. Using the trigonometric formula
sin(a + b) = sin(a) cos(b) + cos(a) sin(b) you get lim (h -> 0) [sin(x) cos(h) + sin(h) cos(x) - sin(x)] / h = (... since the cosine tends to 1...) = lim (h -> 0) sin(h) cos(x)/ h. Now you apply the approximation sin(h) = h for small h and finally find that the derivative is cos(x). This allows you to calculate the Taylor series.

[eranga262154]

Ya, it's true. According to your geomatric derivation can easy who has no well knowledge in derivation.

Anyway thanks.

[krtxmrtz]

Ya, it's true. According to your geomatric derivation can easy who has no well knowledge in derivation.

Anyway thanks.

And I think it's smart, therefore beautiful.

[eranga262154]

Thanks to everyone of you.