Here is a problem I need help doing - once again, an approach would be fine:
What is the minimum possible value of $\cos(\alpha)$ given that, $$ \sin(\alpha)+\sin(\beta)+\sin(\gamma)=1 $$ $$ \cos(\alpha)+\cos(\beta)+\cos(\gamma)=1 $$ The minimum value can be expressed as $-a/b-\sqrt{c}/d$, where $a$ and $b$ are relatively prime, and $c$ isn't divisible by any primes $d$ is divisible by. Find $a+b+c+d$.
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As an alternative, we have that
$$\cos x+\cos y+\cos z=1$$ $$i\sin x+i\sin y+i\sin z=i $$
and summing up by Euler's identity we obtain
$$e^{ix}+e^{iy}+e^{iz}=\sqrt 2 e^{i\frac \pi 4} \iff e^{i\left(x-\frac \pi 4\right)}+ e^{i\left(y-\frac \pi 4\right)}+ e^{i\left(z-\frac \pi 4\right)}=\sqrt 2$$
which, in order to maximize one angle, has the following nice geometrical intepretation
which, by cosine law, leads to
$$(1+1)^2=1^2+(\sqrt 2)^2-2\cdot 1\cdot \sqrt 2 \cdot \cos \left(x-\frac \pi 4\right)$$
$$\iff x-\frac \pi 4 = \arccos\left(-\frac1{2\sqrt 2}\right)$$
and then, by addition formula
$$\cos x= \cos \left( \arccos\left(-\frac1{2\sqrt 2}\right)+\frac \pi 4 \right)=\\=-\frac1{2\sqrt 2}\cdot\frac {\sqrt 2} 2 - \frac {\sqrt 7}{2\sqrt 2} \cdot \frac {\sqrt 2} 2 =\frac{-1-\sqrt{7}}{4}\approx -0.91143\dots$$
Here is a nice Geogebra demonstration of the logic prepared by user Semiclassical.
Rearrange the two given equations,
$$ \sin\beta+\sin\gamma=1-\sin\alpha $$ $$ \cos\beta+\cos\gamma=1 -\cos\alpha$$
and square their both sides,
$$\sin^2\beta+\sin^2\gamma+2\sin\beta\sin\gamma = 1-2\sin\alpha+\sin^2\alpha\tag{1}$$ $$\cos^2\beta+\cos^2\gamma+2\cos\beta\cos\gamma = 1-2\cos\alpha+\cos^2\alpha\tag{2}$$
From (1) + (2),
$$2+2\cos(\beta-\gamma)=3-2(\cos\alpha+\sin\alpha)$$
or,
$$\cos\alpha+\sin\alpha =\frac 12 [1-2\cos(\beta-\gamma)]$$
$\cos\alpha$ takes the minimum value when $\cos(\beta-\gamma)=1$. Thus,
$$\cos\alpha+\sin\alpha =-\frac 12$$
Solve to get,
$$\cos\alpha = \frac{-1-\sqrt{7}}{4}$$
Therefore, $a+b+c+d=16$.