anonymous
  • anonymous
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 ?
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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/k-12/airplane/tablsin.html It should be easy to see from that table that it is never greater than one or less than zero.
anonymous
  • anonymous
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".
anonymous
  • anonymous
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

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