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    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #1

    May 7, 2007, 08:46 AM
    trig proof
    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:

    But I got no idea, what to do next.
    Any help appreciated, thanks.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    May 7, 2007, 11:50 AM
    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.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    May 7, 2007, 01:13 PM
    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?
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #4

    May 7, 2007, 11:32 PM
    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 :).

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