What is the smallest time from the data given

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The discussion centers on determining the smallest time \( t \) from the equation \(\sin(120\pi t)=\frac{2}{5}\). The solution involves the formula \( t=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5} \), where \( n \) is a non-negative integer. The smallest value of \( t \) occurs at \( n=0 \). Further proof of this minimum requires additional mathematical exploration.

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https://www.physicsforums.com/attachments/1729

before I proceed with b, c, and d. what really is meant by the smallest time. does this mean the time of one cycle. there are multiple answers if it calculated for t. the steps I used give one but that may not be the correct process. no ans given
 
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karush said:
https://www.physicsforums.com/attachments/1729

before I proceed with b, c, and d. what really is meant by the smallest time. does this mean the time of one cycle. there are multiple answers if it calculated for t. the steps I used give one but that may not be the correct process. no ans given

Hi karush, :)

You have obtained,

\[\sin(120\pi t)=\frac{2}{5}\]

Generally, the values that satisfies this equation are given by,

\[120\pi t=n\pi+(-1)^n \sin^{-1}\frac{2}{5}\]

where \(n\in \mathbb{Z}^{+}\cup\{0\}\). Therefore we have,

\[ t=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5}\]

So by substituting \(n=0,\,1,\,2\,\cdots\) and so on we can see that at \(n=0\), \(t\) attains it's smallest value. That's not a proof though, I guess to show that \(t(n)=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5}\) has a minimum at \(n=0\) in a more abstract manner will need more work. I can't think of any simple method. :confused:
 

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