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kapilagayan · Jul 30, 2009, 08:51 AM

If A+B+C= pie

Then prove that

cosAcosBcosC=< 1/8

galactus · Jul 30, 2009, 11:45 AM

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:

$sin^{2}(b-c)+cos^{2}(b-c)-4cos(a)cos(b-c)+4(cos^{2}(a))=sin^{2}(b-c)+<br /> (cos(b-c)-2cos(a))^{2}$ .

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: