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Trigonometry What is the smallest time from the data given

karush

Well-known member
Jan 31, 2012
2,770
View attachment 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
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
View attachment 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: