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If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$? Your help is appreciated.I tried various formulas but nothing is working.

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38 minutes. $ $ –  Did Nov 3 '11 at 18:40

4 Answers 4

up vote 2 down vote accepted

$cos x +cos y -cos(x+y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-(2\cos^2(\frac{x+y}{2})-1)=2\cos(\frac{x+y}{2})\cdot\left(\cos(\frac{x-y}{2})-cos(\frac{x+y}{2})\right)+1\le $

$2\cos(\frac{x+y}{2})\cdot\left(1-cos(\frac{x+y}{2})\right)+1\le 1/2+1=\frac{3}{2}$, where the inequality is by $ab\le (\frac{a+b}{2})^2$, so $\cos(\frac{x+y}{2})\cdot\left(1-cos(\frac{x+y}{2})\right)\le 1/4$ . The equality holds iff $x=y=\pi/3$.

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Thank you.The replies were quite instructive. –  Eisen Nov 3 '11 at 18:19

Here's a sketch: Consider $y\in[0,\pi/2]$ as fixed, and let $f(x)=\cos x + \cos y - \cos(x+y)$. Locate the maximum of $f$ on $[0,\pi/2]$ by solving $f'(x)=0$ for $x$; you'll find that the largest value is $$ f\left(\frac{\pi}{2}-\frac{y}{2}\right)=\frac{3}{2}-2 \left(\sin\frac{y}{2}- \frac{1}{2}\right)^2. $$ Maybe you can finish it from there?

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Thank you, but I am not allowed to use calculus. –  Eisen Nov 3 '11 at 17:59
    
Unfortunately, I cannot accept both answers.:) –  Eisen Nov 3 '11 at 18:20

Algebraic but cumbersome: express the condition in terms of $a=\mathrm e^{\mathrm ix}$ and $b=\mathrm e^{\mathrm iy}$ and solve.

One knows that $2\cos(x)=a+a^{-1}$, $2\cos(y)=b+b^{-1}$, and $2\cos(x+y)=ab+a^{-1}b^{-1}$ since $ab=\mathrm e^{\mathrm i(x+y)}$. Multiplying everything by $ab$ to get rid of the denominators, one gets $$ a(1-a)b^2+(a^2+1-3a)b+a-1=0. $$ Solving the second degree equation with unknown $b$ yields $$ b^{-1}=1-a\quad\mbox{or}\quad b=1-a^{-1}. $$ Either $b^{-1}=1-a$, then $|1-a|=1$ hence $a$ is on both circles of radius $1$ centered at $0$ and at $1$, that is $x=\pm\pi/3$, and $y=x$ since $1-a=a^{-1}$.

Or $b=1-a^{-1}$, then $a^{-1}=1-b$ hence the solution is the same than in the first case, exchanging the roles of $x$ and $y$.

Finally, the condition that $x$ and $y$ are in $[0,\pi/2]$ imposes that $x=y=+\pi/3$.

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You could also attempt an geometric proof. First, without loss of generality you can assume $0 <x,y < \frac{\pi}{2}$.

Construct a triangle with angles $x,y, \pi-x-y$.

Let $a,b,c$ be the edges. Then by cos law, you know that

$$\frac{a^2+b^2-c^2}{2ab}+ \frac{a^2+c^2-b^2}{2ac}+ \frac{b^2+c^2-a^2}{2bc}=\frac{3}{2}$$ and you need to show that $a=b=c$. The inequality is

$$c(a^2+b^2-c^2)+b(a^2+c^2-b^2)+a(b^2+c^2-a^2)=3abc$$

or

$$a^2b+ab^2+ac^2+a^2c+bc^2+b^2c=a^3+b^3+c^3+3abc \,.$$

This should be easy to factor, but I fail to see it....

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