Why is this true about the angle theta

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SUMMARY

The discussion centers on the small angle approximation of the sine function, specifically the equation V_{e} \times sin \Theta \cong V_{e} \Theta. The key takeaway is that as the angle Theta approaches zero, the sine of Theta can be approximated by Theta itself due to the Taylor Expansion for sine. This approximation simplifies calculations in physics and engineering when dealing with small angles.

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Pietair
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Homework Statement


From my notes:

Picture2.png


Why is:
[tex]V_{e} \times sin \Theta \cong V_{e}\Theta[/tex] ?

The only thing I know is that the assumption: [tex]\Theta \rightarrow 0[/tex] has been made, but that doesn't make the above equation clear to me...

Thanks in advance!
 
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I'm not 100% sure what you're asking, but it looks like you are just talking about the small angle approximation of the sine function. That's easy to show if you know the Taylor Expansion for sine.

[tex]\sin{\theta} = \theta - {\theta^3 \over 3!} + {\theta^5 \over 5!} - ...[/tex]

Now, since [tex]\theta[/tex] is very small, any positive integer power of [tex]\theta[/tex] will go to 0 even faster than [tex]\theta[/tex] itself. ie. [tex]\theta^3 \to 0[/tex], [tex]\theta^5 \to 0[/tex], etc.

So [tex]\sin{\theta} \approx \theta[/tex]
 


That was exactly my question, the answer is clear, thanks a lot!
 

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