Can SinA > SinB Be Used to Prove A > B in Triangle Angles?

[Michael_Light]

I am in pre-u level, i apologize if i have asked something dumb but it bothers me a lot.

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.

 

[AlephZero]

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.

 

[dalcde]

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.

 

[Hootenanny]

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.

That wouldn't help you in this case since angles in a triangle belong to the set \{\theta:0<\theta<\pi\}.

 

[uart]

I am in pre-u level, i apologize if i have asked something dumb but it bothers me a lot.

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.

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 A = \pi/2 + \theta, then it follows from the angle sum of a triangle that, B < \pi/2 - \theta. So from the symmetry of sin() about pi/2 we can conclude that the inequality sin(A) > sin(B) still holds.

 

[uart]

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 ...

 

[uart]

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 (\sin A>\sin B) and that B is non-acute, say B = \pi/2 + \theta.

Since B = \pi/2 + \theta, then to satisfy the angle sum condition we require A &lt; \pi/2 - \theta. The symmetry of sin() about pi/2 then implies that sin(A) < sin(B), giving us the required contradiction.

 

[HallsofIvy]

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.