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Given 3 non-null vectors $v,u,w$ and angles $a=(u,v), b=(u,w), c=(v,w)$, Prove that $-3/2\leq\cos a + \cos b + \cos c\leq 3$.

I've managed to prove that: $\cos a + \cos b + \cos c\leq 3$ basically arguing that $\cos \theta$ is bounded by $-1,1$ using the inequality of Cauchy-Schwarz. I reasoned that as the maximum value of $\cos x$ is $1$, we could have three angles such that $1+1+1=3$ which is the right answer, but I guess I reasoned incorrectly. And for the lower bound:

$$-3/2\leq\cos a + \cos b + \cos c$$

I have no idea on what to do.

Martin Sleziak
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Red Banana
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3 Answers3

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The picture is basically dividing a circle into 3 sectors and $a,b,c$ are the angles of each sector.

Since, $c=2\pi-a-b$, $\cos c= \cos (a+b)$ and

$$\cos a + \cos b + \cos c = \cos a +\cos b + \cos a \cos b - \sin a \sin b $$

writing $x= \cos a,\; y= \cos b$, and $\sin a =\sqrt{1-x^2} $, etc. (We can always choose $+\sqrt{}$ by choosing the angles to be <$\pi$) We have, $$ x+y+xy -\sqrt{1-x^2}\sqrt{1-y^2}, \quad x,y\in[-1,1] $$

By A.M. > G.M., $$-\sqrt{1-x^2}\sqrt{1-y^2} \ge -\frac{2-x^2-y^2}{2} =-1 +\frac{x^2+y^2}{2}$$ Thus, \begin{align} x+y+xy -\sqrt{1-x^2}\sqrt{1-y^2} & \ge x+y+xy-1+ \frac{x^2+y^2}{2}\\ &= \frac{1}{2}(x+y+1)^2 -\frac{3}{2} \ge -\frac{3}{2}. \end{align} And equality holds if and only if $x=y=-1/2$, which means the angles are all $ \frac{2\pi}{3}$.

mastrok
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Hint: Either we have $a+b=c$ (or permutation) or $a+b=2\pi-c$.

Quang Hoang
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If a = b = c = 2 * pi / 3, then cos a + cos b + cos c = -3/2. This looks like the only way to obtain -3/2 - both in 2D- and in 3D-cases.

Sergei M.
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