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 $A < \pi/2 - \theta$. 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.