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If $A,B,C$ angles of a triangle aka $A+B+C=180°$, and I need to evaluate $$\cos(A)+\cos(B)+\cos(C)$$ I'm out to get that at most it is $$\cos(A)+\cos(B)+\cos(C)<=3/2$$ Since: $A+B+C=180°$, then $(A+B)/2=(180°-C)/2=90°-C/2$

I can shown that: $$\cos(A)+\cos(B) = 2cos((A+B)/2)cos((A-B)/2)=2cos(90° -C/2)cos((A-B)/2)$$

Next in the proof comes the bit I do not get - the evaluation: $$2cos((A+B)/2)cos((A-B)/2)<2cos(90° -C/2)$$

Why can this be done? Is there some inequality that presents that?? Or does it come simply from the fact that cosine function is bound between [-1;1] and that taking away $$*cos((A-B)/2$$ will make $$2cos(90° -C/2)cos((A-B)/2)$$ bigger?

The end result is what I need: $$2cos((A+B)/2)cos((A-B)/2)<2cos(90° -C/2)=2sin(c/2)$$ $$\cos(A)+\cos(B)+\cos(C)<=2sin(c/2)+\cos(C)=2sin(c/2)+1-2sin^2(c/2)=-2(2sin(c/2)-1/2)^2 +3/2<=3/2$$

Den
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  • Your question will get more attention if it is well-formatted, use consistent degrees/radians and is explained in a precise manner. – Shailesh Jan 14 '16 at 17:49
  • There seem to be some unbalanced parantheses too. – Shailesh Jan 14 '16 at 17:57
  • You have to edit because if $A=B=C=60^{\circ}$ then $1<1$. It is maybe that factor $2$ in both sides. I guess it is mainly with $\le$ instead of $<$ – Piquito Jan 14 '16 at 19:19
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    @Den, See http://math.stackexchange.com/questions/949530/in-triangle-abc-show-that-1-lt-cos-a-cos-b-cos-c-le-frac-32 – lab bhattacharjee Jan 15 '16 at 06:23

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