Trigonometry Proof: For the sharp angles a and b is: sin (a + b) < sin a + sin b

[derf789]

Hi everyone!
I am new here and I also just started to solve mathematical proofs and some of this proofs really confuse me. My teacher gave me some proofs to do voluntarily and i solved the first few but they were easier than this one:

The original text is in german but i try to translate it as best as possible

Proof that this Statement is correct:

For the sharp angles a and b is: sin (a + b) < sin a + sin b.

Note: 0<cos a< 1

I really dont know where to start and i really would appreciate any help!

 

[Dr.Peterson]

Hi everyone!
I am new here and I also just started to solve mathematical proofs and some of this proofs really confuse me. My teacher gave me some proofs to do voluntarily and i solved the first few but they were easier than this one:

The original text is in german but i try to translate it as best as possible

Proof that this Statement is correct:

For the sharp angles a and b is: sin (a + b) < sin a + sin b.

Note: 0<cos a< 1

I really dont know where to start and i really would appreciate any help!

First, I suppose a more standard translation would be,

For any two acute angles a and b, sin(a + b) < sin(a) + sin(b).

Now, it will help if you can tell us how much trigonometry you know. In particular, do you know the "angle sum formula" with which you can express sin(a + b) in terms of functions of a and b? If so, then you can use that here.

 

[JeffM]

There is no certain way to find a proof. One method is to just start by trying things that have already been shown true.

In this case, just start by remembering that

\(\displaystyle sin(x + y) = sin(x) * cos(y) + sin(y) * cos(x)\)

and by remembering that \(\displaystyle z + 0 = z = z * 1.\)

The first point is basic trigonometry. The second point is basic arithmetic and comes up in lots of proofs.

Now can I show that

\(\displaystyle sin(x) * cos(y) + sin(y) * cos(x) < sin(x) + sin(y) = sin(x) * 1 + sin(y) * 1.\)

That would be true if both sines > 0 and both cosines < 1. BUT THEY ARE.

So now it is just a matter of proper ordering and notation. You figure out the reasoning any way you can before you figure out a pretty sequence.

\(\displaystyle \text {GIVEN: } 0 < x < \dfrac{\pi}{2} \text { and } 0 < y < \dfrac{\pi}{2}.\)

\(\displaystyle \therefore 0 < sin(x) \text { and } cos(y) < 1 \implies sin(x) * cos(x) < sin(x) * 1 = sin(x).\)

\(\displaystyle \text {And } 0 < sin (y) \text { and } cos(x) < 1 \implies sin(y) * cos(x) < sin(y) * 1 = sin(y).\)

\(\displaystyle \therefore sin(x) * cos(y) + sin(y) * cos(x) < sin(x) + sin(y).\)

\(\displaystyle \text {BUT } sin(x + y) = sin(x) * cos(y) + sin(y) * cos(x).\)

\(\displaystyle \therefore sin(x + y) < sin(x) + sin(y). \text { Q.E.D.}\)