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If sinA > sinB, then A>B?

  1. Sep 19, 2011 #1
    I am in pre-u level, i apologize if i have asked something dumb but it bothers me a lot. :frown:

    Here it goes my question: If A and B are angles inside a triangle, and sinA > sinB, can i conclude that A>B? If yes, how can we prove it? Can i use this argument in proving question?

    Please guide me. :smile:
     
  2. jcsd
  3. Sep 19, 2011 #2

    AlephZero

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    Your restriction to "angles inside a triangle" means angles between 0 and 180 degrees (or 0 and pi radians).

    Look at a graph of the sin function between 0 and 180 degrees, and you should be able to see why this is false.

    If you restrict the range to between 0 and 90 degrees not 180, then it is true (that is also clear from looking at the graph).

    Of course looking at a graph isn't really a "proof", but proving this rigorously needs math beyond what you have probably studied at pre-university level.
     
  4. Sep 19, 2011 #3
    Can't we just find the derivative of sin A and see that it is always positive in (0,90), hence strictly increasing. That's something you have to study pre-university in Hong Kong.
     
  5. Sep 19, 2011 #4

    Hootenanny

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    That wouldn't help you in this case since angles in a triangle belong to the set [itex]\{\theta:0<\theta<\pi\}[/itex].
     
  6. Sep 19, 2011 #5

    uart

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    Yes, provided that A and B are angles within the same triangle, then A > B implies that sin(A) > sin(B).

    It's easiest to consider the proof as two cases, one for A acute and another A obtuse.

    Case 1. If the larger angle "A" is less than or equal pi/2 then the proof is obvious, since sin is an increasing function on [0..pi/2).

    Case 2. If the larger angle "A" is greater then pi/2, say [itex]A = \pi/2 + \theta[/itex], then it follows from the angle sum of a triangle that, [itex]B < \pi/2 - \theta[/itex]. So from the symmetry of sin() about pi/2 we can conclude that the inequality sin(A) > sin(B) still holds.
     
    Last edited: Sep 19, 2011
  7. Sep 19, 2011 #6

    uart

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    Sorry I misread your question as :
    - "does A > B in a triangle imply that sin(A) > sin(B)"

    when you actually asked :
    - "does sin(A) > sin(B) in a triangle imply that A > B"

    Let me think for a minute if the implication goes both ways ...
     
    Last edited: Sep 19, 2011
  8. Sep 19, 2011 #7

    uart

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    Ok, the implication does indeed work the other way too. One way to prove this is to start by proving the following:

    - if sinA > sinB (in a triangle) then B must be acute. After proving this, then the result you seek is very straight forward.

    One way to prove the result about "B" being acute is by contradiction. Assume that [itex](\sin A>\sin B)[/itex] and that B is non-acute, say [itex]B = \pi/2 + \theta[/itex].

    Since [itex]B = \pi/2 + \theta[/itex], then to satisfy the angle sum condition we require [itex]A < \pi/2 - \theta[/itex]. The symmetry of sin() about pi/2 then implies that sin(A) < sin(B), giving us the required contradiction.
     
    Last edited: Sep 19, 2011
  9. Sep 20, 2011 #8

    HallsofIvy

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    Regarding AlephZero's response, it is true that just "sin(A)> sin(B)" does NOT imply "A> B", even if A and B are restricted to be between 0 and 180 degrees. However, for A and B two angles in a triangle, it is a different matter. It is true that if A> 90- x, B> 90+ x, then sin(A)> sin(B) while A> B. However, in that case, we have A+ B> 180 degrees so they cannot be two angles in a triangle.
     
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