Given
$$\begin{align}
\sin x+\sin y+\sin z &=2 \\[4pt]
\cos x+\cos y+\cos z &=\frac{11}{5} \\[4pt]
\tan x+\tan y+\tan z &=\frac{17}{6}
\end{align}$$where $x,y,z\in\mathbb{R}.$ Find the value of $\sin{(x+y+z)}$, without a calculator.
By "without a calculator", I mean without any electronic computing device, i.e. just pen and paper. I do not know if such a solution is possible.
I made up this problem. The answer is $4/5$, but I don't know how to find it without a calculator.
When making the problem, to get nice numbers in the question and answer, I let $x=y=\arctan\frac34$ and $z=\arctan\frac43$.
My attempt:
Let $A=\cos x, \space B=\cos y, \space C=\cos z$.
Since $\sin x+\sin y+\sin z=2$, we know that $\sin x$, $\sin y$, $\sin z$ are all non-negative. So we get the following two equations with $A$ and $B$:
$$\sqrt{1-A^2}+\sqrt{1-B^2}+\sqrt{1-(2.2-A-B)^2}=2$$
$$\frac{\sqrt{1-A^2}}{A}+\frac{\sqrt{1-B^2}}{B}+\frac{\sqrt{1-(2.2-A-B)^2}}{2.2-A-B}=\frac{17}{6}$$
But I don't know how to solve for $A$ and $B$, nor $\sin{(x+y+z)}$.
I also tried, without success, to use the identities
$$(\sin x+\sin y+\sin z)^2+(\cos x+\cos y+\cos z)^2$$
$$=3+2(\cos{(x-y)}+\cos{(y-z)}+\cos{(z-x)})$$
$$\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
I also tried, without success, to use complex numbers.
Best Answer
Let $x=e^{ix}$, $y=e^{iy}$, $z=e^{iz}$. We want to find $\text{Im}(xyz)$.
We can express this system of equations as
$$x+ y+z=\omega :=\frac{11}{5}+2i,$$
and
$$\left(x - \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right) + \left(x + \frac{1}{x}\right)\left(y - \frac{1}{y}\right)\left(z + \frac{1}{z}\right) +\left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z - \frac{1}{z}\right) = \frac{17i}{6}\left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right),$$
which after expanding and simplifying, gives
$$(18-17i)x y z +(6-17i)\left(\frac{x y}{z} + \frac{x z}{y}+ \frac{y z}{x}\right) -(6+17i)\left(\frac{x}{yz} + \frac{y}{xz}+ \frac{z }{xy}\right) -\frac{18+17i}{xyz} = 0.$$
Noting the symmetry, this implies
$$\text{Im}\left((18-17i)x y z +(6-17i)\underbrace{\left(\frac{x y}{z} + \frac{x z}{y}+ \frac{y z}{x}\right)}_{A}\right) = 0.$$
The plan is to express $A$ in terms of $xyz$, then find solutions for $xyz$.
Taking the conjugate, the first condition implies that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\overline \omega.$$
Squaring, we have
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{xz}+\frac{2}{yz}=\overline\omega^2,$$
which gives for $A$
$$\frac{x y}{z} + \frac{x z}{y}+ \frac{y z}{x} = \overline\omega^2 xyz-2\omega.$$
Substituting, we have $$\text{Im}\left((18-17i)x y z +(6-17i)\left(\overline\omega^2 xyz-2\omega\right)\right) = 0,$$
or
$$\text{Im}\left(\left(-\frac{3164}{25}-\frac{2102}{25}i\right)x y z -\frac{472}{5}+\frac{254}{5}i\right)=0.$$
This implies
$$1051\text{Re}(xyz)+1582\text{Im}(xyz) = 635.$$
The only solutions to
\begin{equation} \begin{cases} 1051 a + 1582b =635\\ a^2+b^2=1 \end{cases} \end{equation}
are $\left(-\frac{3}{5}, \frac{4}{5}\right)$ and $\left(\frac{699833}{721465},-\frac{175344}{721465}\right)$. However, as other solutions have already pointed out, we know all the angles are in the first quadrant, so the second solution is extraneous. We conclude that $\text{Im}(xyz)=\frac{4}{5}$.
This last step is admittedly tedious, but potentially doable by hand: using the unit circle parameterization $t\mapsto\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$, this boils down to solving $843 t^2 -1582t -208=0.$ Either notice that this factors as $(t-2)(843t+104)$, or use the quadratic formula to get $t = \frac{1582\pm\sqrt{3204100}}{2(843)} = \frac{1582\pm 1790}{2(843)}$.