# I Vector math (small angle approximation)

#### Boltzman Oscillation

Given the following vectors:

how can i determine that Θ = Δp/p ?
I can understand that p + Δp = p' but nothing arrives from this. Any help is welcome!

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#### anorlunda

Mentor
Sin Φ ≈ ∅ for small angles.

(Different) Mentor edit: The above should be $\sin(\theta) \approx \theta$, for $\theta$ in radians.

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#### Delta2

Homework Helper
Gold Member
Sin Φ ≈ ∅ for small angles.
You meant to say that $\sin\theta\approx \theta$ for small angles.

#### Boltzman Oscillation

Sin Φ ≈ ∅ for small angles.
wait wait what? How is this true? I've never known this!!

#### Boltzman Oscillation

You meant to say that $\sin\theta\approx \theta$ for small angles.
Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.

#### jedishrfu

Mentor
To be clear, the angle is in radian measure.

#### A.T.

Science Advisor
Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.
You also see that from the fact that sin(0) = 0, sin'(0) = 1 and sin''(0) = 0. So sin is equivalent to the identity function up to the 2nd derivative around 0.

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#### RPinPA

Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.
It's instructive to take $\sin(\theta)$ for some small values of $\theta$ (always in radians) to see just how good the approximation is. Even at $\theta = 0.1$, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.

#### sophiecentaur

Science Advisor
Gold Member
how can i determine that Θ = Δp/p ?
Are you looking for more than just basic Trigonometry here?

#### Boltzman Oscillation

Are you looking for more than just basic Trigonometry here?
Well the other gentlemen helped me understand better now but maybe you can provide more insight. So, yes.

#### Boltzman Oscillation

It's instructive to take $\sin(\theta)$ for some small values of $\theta$ (always in radians) to see just how good the approximation is. Even at $\theta = 0.1$, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.
okay i will try some small numbers. Thank you sir.

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