If $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then how are $x$ and $y$ related? 
If $$\cos x + \cos y - \cos(x+y) = \frac{3}{2}$$ then
  
  
*
  
*(a) $x + y = 0$
  
*(b) $x = 2 y$
  
*(c) $x = y$
  
*(d) $2 x = y$
  

It is problem of trigonometry, and I have the solution of the problem. However, after seeing the solution, I don't quite understand how is one supposed to know how to approach this problem. My request is:

Don't just solve the problem, but also tell me from where do you come to know which approach would work. (Tell me from where do you come to know which trick or approach to be used to solution this problem.)

Thanks in advance.
 A: For a fixed value of $y\in[-\pi,\pi]$, let we consider the stationary points of the function
$$ f(x)=\cos(x)+\cos(y)-\cos(x+y)-\frac{3}{2}.$$
Since $f'(x) = -\sin(x)+\sin(x+y)$, $\,f(x)$ attains its maximum and minumum value over $[-\pi,\pi]$ at $x=\frac{\pi-y}{2}$ and $x=-\frac{\pi+y}{2}$. In particular
$$ f\left(-\frac{\pi+y}{2}\right) = -\frac{3}{2}+\cos(y)-2\sin\left(\frac{y}{2}\right) $$
$$ f\left(\frac{\pi-y}{2}\right) = -\frac{3}{2}+\cos(y)+2\sin\left(\frac{y}{2}\right) $$
and $f$ is bounded between these values. In order that $f(x)=0$ has some solution, the product of such values has to be $\leq 0$, that is equivalent to:
$$ \frac{1}{4}\left(1-2\cos y\right)^2 \color{red}{\leq} 0 $$
so, in order that $f(x)=0$ has some solution, $y$ has to be $\pm\frac{\pi}{3}$, and the associated solution is a stationary point. Now it is straightforward to check that $\color{red}{(c)}$ is the correct answer.

Obviously this solution requires time. In a test, assuming that there actually is a correct answer, it is way faster to notice that $(b)$ and $(d)$ are equivalent and $(a)$ is absurd, hence $(c)$ is the only plausible option. Nitpickers may be happy to notice that $x=\frac{7\pi}{3},y=\frac{\pi}{3}$ is a solution that violates all the given options!
A: Let $z_1 = -\cos x + i\sin x$, $z_2 = -\cos y + i\sin y$. Then,
\begin{align*}
|1+z_1+z_2|^2 \geq 0 &\Rightarrow 3 -2\cos x - 2\cos y + 2\cos(x+y) \geq 0\\
&\Rightarrow \cos x + \cos y - \cos(x+y) \leq \frac{3}{2}
\end{align*}
Equality holds if and only if $1 + z_1 + z_2 = 0$ and hence $1+\bar{z_1}+\bar{z_2} =0$. Adding, we obtain, $\cos x + \cos y = 1$ and consequently $\cos(x+y) = -\frac{1}{2}$. Thus $x+y = \frac{2\pi}{3}$ and from $\cos x + \cos y = 1 $, we get $x = \frac{\pi}{3}$ and $y = \frac{\pi}{3}$. Thus $x=y$ 
This can also be solved without using complex numbers:
\begin{align*}
&\qquad \cos x + \cos y - \cos(x+y) = \frac{3}{2} \\
&\Rightarrow 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) - 2\cos^2\left(\frac{x+y}{2}\right) + 1 = \frac{3}{2} \\
&\Rightarrow 4\cos^2\left(\frac{x+y}{2}\right)  - 4\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) +1 = 0 \\
&\Rightarrow \left(2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right)\right)^2 + 1 - \cos^2\left(\frac{x-y}{2}\right) = 0\\
&\Rightarrow \left(2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right)\right)^2  + \sin^2\left(\frac{x-y}{2}\right)  = 0
\end{align*}
and hence
\begin{align*}
\sin\left(\frac{x-y}{2}\right) &= 0\\
2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right) &= 0
\end{align*}
Thus $x=y = \dfrac{\pi}{3}$
A: Answer B) and D) are essentially the same due to symmetry.
For answer A) if we plug in $x+y=0$ in $cos(x+y)$ we get $cosx+cosy=5/2$ which cannot have a real solution since the sum of these cosines cannot exceed $2$. That leaves us with option C)
A: HINT...If only one answer can be true you can eliminate b and d since they are both the same.
You can eliminate a automatically...
A: Using Prosthaphaeresis Formula,
$$\cos x+\cos y=2\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
Double angle formula says:
$$\cos(x+y)=2\cos^2\dfrac{x+y}2-1$$
So we have $$2\cos^2\dfrac{x+y}2-2\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\dfrac12=0\ \ \ \  (1)$$ which is a Quadratic Equation in $\cos\dfrac{x+y}2$ which is real,
so the discriminant $\left(2\cos\dfrac{x-y}2\right)^2-4=-4\sin^2\dfrac{x-y}2$  must be $\ge0$
So we must have $\sin^2\dfrac{x-y}2=0\implies\dfrac{x-y}2\equiv0\pmod\pi\implies x\equiv y\pmod{2\pi}$
Consequently, $(1)$ becomes,  $$2\cos^2\dfrac{x+y}2-2\cos\dfrac{x+y}2+\dfrac12=0\ \ \ \  (2)\implies\cos\dfrac{x+y}2=\dfrac12$$
$$\implies\dfrac{x+y}2=2m\pi\pm\dfrac\pi3\iff\cdots$$
 where $m$ is any integer
Can you take it from here?
