Could you try to show that sin(180-C) = sin(C)? If that's true, then you're pretty much done, right? If it's false, then the original claim is false.
In any triangle sin(A + B)= sinC
State true or false and explain why.
The angles A,Band C are alpha, betta and gamma I just dont know how to put that in there so I used A,B andC.
I tried to use an addition formula for sin and expand A+B to sines and cosines but I dont think that is right. The teacher wrote on the paper A+B=180-C which makes sense but there must be more than just that. If someone could get me going in the right direction I would appreciate it.
Thanks for the quick reply. So the statement is true. At first I thought that angles A and B would have to be acute to add up to angle C but that does not seem to be the case. You can take the sine of any two angles and add them and as long as the third angle is the supplement. Why is this? Does it have to do with the unit circle and sine being positive from 0 to 180. Also they what you to explain why. How do I put it into words?
No, you cannot "prove by example", which is what you're proposing. You must prove in the abstract. Examples are useful in proofs only insofar as they make a proposition believable, but you can never prove by example (at least, not in this case. There is the proof method of "proof by cases", but that's only valid if you can enumerate every single example and prove all of them). You must show that sin(180-C) = sin(C) for all C. Considering that C can be any real number, you're going to be at it a very long time if you try to prove by example.
Instead, I would recommend that you look at some trig identities, especially sum and difference identities. What ideas does that give you?
I agree with dwsmith. I'd say you're pretty much done. All you have to do, really, is preface your post # 7 with three lines:
A + B + C = 180, which implies
A + B = 180 - C, which implies
sin(A+B) = sin (180-C),
and then you insert your Post # 7 here. Got it?