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hi! i hope that somebody could help me.
i can't understand why 0 < sin x< 1 on (0,π)
if i put π/2 i get 1. could someone help me ?
 one year ago
 one year ago
hi! i hope that somebody could help me. i can't understand why 0 < sin x< 1 on (0,π) if i put π/2 i get 1. could someone help me ?
 one year ago
 one year ago

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acronosBest ResponseYou've already chosen the best response.0
I’m not a math major, but I think I know the answer to this. Sine is modeled using a unit circle. That’s basically a circle drawn on an X and Y plane with the origin at (0,0) and a radius of 1. Since the radius is one, the largest value of x is when y=0 and the largest value of y is when x=0. You can see a unit circle and they talk about the nature of Sine here: http://www.youtube.com/watch?v=JPWVnpdfa0E We are primarily interested in the y axis since the will be opposite in the Sine formula of opposite/hypotenuse. From that graph, it’s pretty easy to see that the value of y is going to never be greater than 1 and never less than 0 for any angle between 0 and pi. Also, since y is always going to be less than the radius, it will be smaller than the hypotenuse which is the radius. A smaller number(x) divided by a bigger number 1(the radius) will be a fraction that is less than 1. However, it is possible for sin to be negative if the angles are greater than pi. At that point, y becomes negative which changes the sign of the result. The original problem was restricted between 0 and pi though, so that will not happen in the problem you stated. You can see this in a table of sin values here: http://www.grc.nasa.gov/WWW/k12/airplane/tablsin.html It should be easy to see from that table that it is never greater than one or less than zero.
 one year ago

acronosBest ResponseYou've already chosen the best response.0
Oops. Disregard my previous post. I didn't read your question carefully enough. I think you are correct. The problem is wrong. It should be "0<= sin x<=1".
 one year ago

BrmathmajorBest ResponseYou've already chosen the best response.2
I believe the problem should state 0 < sinx <= 1 on (0,Pi). If you look at intervals the way we do then (0,Pi) is an open interval so you would never actually use 0 or Pi so the 0<Pi is ok.
 one year ago
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