Hey, how do I prove this inequality?
sin(a+b)<sin a + sin b
If I write out the sin(a+b) I get this:
sin a cos b + sin b cos a - sin a - sin b<0
But I got no idea, what to do next.
Any help appreciated, thanks.

Start with sin(a+b) - sin(a)* cos(b) + sin(b)* cos(a). What's the maximum value the cosine of an angle can be? And from that, then how does sin(a)*cos(b) compare to sin(a)? You're almost home.

One correction to the problem statement however -- I believe that sin(a+b) is less than or equal to sin(a) + sin(b). Check the case where a = b = 0.

Actually, upon reflection, I now realize that if angles a and/or b are in the 3rd or 4th quadrant , then sin(a+b) may be greater than sin(a) + sin(b) - well actually, less negative. Which makes me wonder whether the original question either was restricted to angles and b in the first quadrant, or had absolute value symbols that had been removed? Kristo?

Yes, the angles were restricted to the first quadrant, forgot to add it in the first post, sorry!
Thanks for your help, I got it now :).