If A+B+C= pie

Then prove that

cosAcosBcosC=< 1/8

If a+b+c= {\pi}, then prove that cos(a)cos(b)cos(c)\leq \frac{1}{8}



That is Pi, not Pie:rolleyes: This is math, not a bakery. Pie is a tasty treat. Pi is the 16th letter of the Greek alphabet used to represent various mathematical properties, namely, the transcendental number that denotes the ratio of a circles circumference to its diameter. :D

Anyway, we can use a product to sum formula for this.

1 - 8cos(a)cos(b)cos(c) = 1-4cos(a)(cos(b-c)+cos (b+c)).

Now, note that cos(b+c)=-cos({\pi}-b-c)=-cos(a).

So, we get:

sin^{2}(b-c)+cos^{2}(b-c)-4cos(a)cos(b-c)+4(cos^{2}(a))=sin^{2}(b-c)+
(cos(b-c)-2cos(a))^{2}.

But sin^{2}(b-c)+cos^{2}(b-c)=1.

1-8cos(a)cos(b)cos(c)\geq 0

-8cos(a)cos(b)cos(c)\geq -1

Dividing by -8 reverses the equality sign:

cos(a)cos(b)cos(c)\leq \frac{1}{8}

I would think one could also do this with calculus.


I hope I didn't make any typos in all that conglomeration:eek: