# Trig proof question

1. May 17, 2005

### Kamataat

Let $\alpha,\beta\in]0;\pi/2[$. Prove that $\sin(\alpha+\beta)<\sin\alpha+\sin\beta$.

My intuition says it's true, as can be seen from $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$, because $0<\cos\alpha,\cos\beta<1$, but I haven't been able to prove it.

- Kamataat

2. May 17, 2005

### Curious3141

The inequality should not be a strict one, it should include the "=" sign for the domain given. But otherwise, you're completely on the right track and you've already proven it. You've already made the observation that the cosines are positive for the domain and have magnitude less than or equal to one.

$$\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \leq \sin\alpha + \cos\alpha \sin\beta \leq \sin\alpha + \sin\beta$$

and you're done. Note that doing it in two "stages" makes things easier to see. The equality occurs when $\alpha = \beta = 0$.

EDIT : I should mention that the fact that both sines are also non-negative over the domain is critical in the proof.

Last edited: May 17, 2005
3. May 17, 2005

### whozum

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\ sin\beta$

Perhaps the identity $sin(x) = cos(\frac{\pi}{2} - x)$ would be helpful. It goes both ways.

$\sin(\alpha+\beta)=\sin\alpha\sin(\frac{\pi}{2} - \beta)+\sin(\frac{\pi}{2} - \alpha) sin\beta$

Then you get

$$\sin\alpha\sin\left(\frac{\pi}{2} - \beta\right)+\sin\left(\frac{\pi}{2} - \alpha\right) sin\beta < \sin\alpha + \sin\beta$$

Unsure thoughts :
For $\alpha, \beta > \frac{\pi}{4} [/tex] you are bounded underneath by [itex] \sqrt{2} [/tex] on the RHS, but the left will be bounded underneath by 1. Since sine is increasing between [itex] \frac{\pi}{4} \rightarrow \frac{\pi}{2} [/tex] it holds for that interval. Perhaps the easiest way would just be to show that [itex]0 <\sin\alpha+\sin\beta - \sin(\alpha+\beta)$ by saying that in the given domain the function is bounded below by 0, and is increasing throughout?