If $\sin x+\sin y+\sin z=2$, $\cos x+\cos y+\cos z=11/5$, $\tan x+\tan y+\tan z=17/6$, $x,y,z\in\mathbb{R},$ find $\sin(x+y+z)$ without a calculator

Question

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.

Answer 1

Remarks: Here is a human verifiable solution. The calculation (10)-(11) is not easy by hand.

Problem: Let $x, y, z$ be real numbers such that \begin{align*} \sin x + \sin y + \sin z &= 2, \tag{1}\\ \cos x + \cos y + \cos z &= \frac{11}{5}, \tag{2}\\ \tan x + \tan y + \tan z &= \frac{17}{6}. \tag{3} \end{align*} Find the value of $\sin (x + y + z)$.

Solution.

Due to symmetry, assume that $\cos z = \min(\cos x, \cos y, \cos z)$. Thus, $\cos z \le \frac{11}{15}$.

From (2), we have $\cos x, \cos y, \cos z \ge 1/5$. From (1), we have $\sin x, \sin y, \sin z > 0$.

From (1) and (2), using the identity $\frac{\sin x + \sin y}{\cos x + \cos y} = \tan \frac{x + y}{2}$, we have $$\tan \frac{x + y}{2} = \frac{2 - \sin z}{11/5 - \cos z} \tag{4}$$ which results in \begin{align*} \cos (x + y) &= \frac{1 - \tan^2 \frac{x + y}{2}}{1 + \tan^2 \frac{x + y}{2}} = \frac{25\cos^2 z + 50\sin z - 55\cos z - 2}{123 - 50\sin z - 55\cos z}, \tag{5}\\ \quad \sin(x + y) &= \frac{2\tan \frac{x + y}{2}}{1 + \tan^2 \frac{x + y}{2}} = \frac{25\sin z\, \cos z - 55\sin z - 50\cos z + 110}{123 - 50\sin z - 55\cos z}. \tag{6} \end{align*}

From (1) and (2), we have $$(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = (2 - \sin z)^2 + (11/5 - \cos z)^2 \tag{7}$$ which results in $$2\cos (x - y) = (2 - \sin z)^2 + (11/5 - \cos z)^2 - 2. \tag{8}$$

Using (5) and (6), we have \begin{align*} \tan x + \tan y &= \frac{2\sin (x + y)}{\cos (x + y) + \cos (x - y)}\\ &= 2\cdot \frac{\frac{25\sin z\, \cos z - 55\sin z - 50\cos z + 110}{123 - 50\sin z - 55\cos z}}{\frac{25\cos^2 z + 50\sin z - 55\cos z - 2}{123 - 50\sin z - 55\cos z} + \frac{ (2 - \sin z)^2 + (11/5 - \cos z)^2 - 2}{2}}. \tag{9} \end{align*}

From (3) and (9), we have \begin{align*} 2\cdot \frac{\frac{25\sin z\, \cos z - 55\sin z - 50\cos z + 110}{123 - 50\sin z - 55\cos z}}{\frac{25\cos^2 z + 50\sin z - 55\cos z - 2}{123 - 50\sin z - 55\cos z} + \frac{ (2 - \sin z)^2 + (11/5 - \cos z)^2 - 2}{2}} + \frac{\sin z}{\cos z} = \frac{17}{6}. \tag{10} \end{align*}

Let $s = \sin z$ and $c = \cos z$. From (10), using $c^2 + s^2 = 1$, we have $$ \left( -158200\,{c}^{2}+137840\,c+174048 \right) s = 105100\,{c}^{3} - 547620\,{c}^{2} + 361136\,c + 117600$$ or (squaring both sides) \begin{align*} &\left( -158200\,{c}^{2}+137840\,c+174048 \right)^2(1 - c^2)\\ ={}& (105100\,{c}^{3} - 547620\,{c}^{2} + 361136\,c + 117600)^2 \end{align*} or \begin{align*} &-16\, \left( 5\,c-3 \right) \left( 5\,c-4 \right) ^{2}\\ &\times \left( 18036625\,{c}^{3}-39680575\,{c}^{2 }+41195280\,c+21436128 \right) = 0. \tag{11} \end{align*}

Using $1/5 \le c \le \frac{11}{15}$, from (11), we have $c = 3/5$. Then $s = \sqrt{1 - c^2} = 4/5$.

Thus, using (5) and (6), we have \begin{align*} \sin (x + y + z) &= \sin(x + y)\, \cos z + \cos(x + y)\, \sin z \\ &= \frac45. \end{align*}

We are done.

Answer 2

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 $\text{Im}(xyz)=\frac{4}{5}$.

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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 $t = \frac{1582\pm\sqrt{3204100}}{2(843)} = \frac{1582\pm 1790}{2(843)}$.

Answer 3

Don't know if you did (or had reasons to) expect a "nice" solution to this. For what it's worth, here is the brute force solution.

Let $\,\sin x=u\,$, $\,\sin y=v\,$, $\,\sin z=w\,$, $\,\cos x=a\,$, $\,\cos y=b\,$, $\,\cos z=c\,$, then $\,\sin(x+y+z) = t\,$ is the solution of a quadratic equation that results by eliminating $\,u,v,w,a,b,c\,$ between the following:

$$ \begin{cases} \begin{align} u+v+w &= 2 \\ a+b+c &= \frac{11}{5} \\ \frac{u}{a}+\frac{v}{b}+\frac{w}{c}&=\frac{17}{6} \\ u^2 + a^2 &= 1 \\ v^2 + b^2 &= 1 \\ w^2 + c^2 &= 1 \\ t &= ubc+vca+wab-uvw \end{align} \end{cases} $$

Courtesy WA, the equation in $t$ returned by GroebnerBasis[{u+v+w-2, 5(a+b+c)-11, 6(ubc+vca+wab)-17abc, u^2+a^2-1, v^2+b^2-1, w^2+c^2-1, t-(ubc+vca+wab-uvw)}, {t}, {u,v,w,a,b,c}] is:

$$ 3607325 t^2 - 2009140 t - 701376 = 0 $$

The solutions are $\,t \in \left\{-\frac{175344}{721465}, \frac{4}{5}\right\}\,$.

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[ EDIT ] $\;$ The above was posted before the restriction to real solutions $\,x,y,z \in \mathbb R\,$ was added to the question. With that restriction, the only real solution is indeed $\,t = \frac{4}{5}\,$ since the negative $t$ corresponds to a solution with two non-real complex values. This can be verified by solving the system for $\,\sin x = a\,$, for example, which gives $\,a \in \left\{\frac{3}{5}, \frac{4}{5}\, -0.368\right\}\,$. The positive solutions are permutations of the real solution with $\,t = \frac{4}{5}\,$. The negative solution gives $\,\sin y + \sin z = 2.368\,$ which means $\,y,z \in \mathbb C \setminus \mathbb R\,$.

Answer 4

The rational numbers led me to wonder if all nine trig values might be rational. That would require each of the three angles to be an angle in a Pythagorean triple triangle.

Then the cosines summing to something with denominator $5$ leads me to consider a 3-4-5 triangle, where one angle has cosine $\frac35$ and another has cosine $\frac45$. And then you ask if several of these could sum to $\frac{11}5$. Indeed: $$\frac45+\frac45+\frac35=\frac{11}5$$

And sure enough, with the corresponding sines: $$\frac35+\frac35+\frac45=\frac{10}5=2$$

And with the corresponding tangents: $$\frac34+\frac34+\frac43=\frac{17}6$$

So one specific solution is $x=\arcsin(3/5)$, $y=\arcsin(3/5)$, $z=\arcsin(4/5)$. (Well, three solutions for the three permutations of this one solution.) For this one solution you can work out what is $\sin(x+y+z)$ and of course it works out to $\frac45$:

$$\begin{align} &\sin(2\arcsin(3/5)+\arcsin(4/5))\\ &=\sin(2\arcsin(3/5))\cos(\arcsin(4/5))+\cos(2\arcsin(3/5))\sin(\arcsin(4/5))\\ &=\sin(2\arcsin(3/5))\cdot\frac35+\cos(2\arcsin(3/5))\cdot\frac45\\ &=2\sin(\arcsin(3/5))\cos(\arcsin(3/5))\cdot\frac35+\left(\cos^2(\arcsin(3/5))-\sin^2(\arcsin(3/5))\right)\cdot\frac45\\ &=2\left(\frac35\right)\left(\frac45\right)\cdot\frac35+\left(\left(\frac45\right)^2-\left(\frac35\right)^2\right)\cdot\frac45=\frac{72}{125}+\frac{28}{125}=\frac{100}{125}=\frac45 \end{align}$$

This leaves open the possibility of more solutions and $\sin(x+y+z)$ not being constant among those solutions. But if you know ahead of time that $\sin(x+y+z)$ is constant...

Answer 5

Best use complex numbers - as in the answer by user51547 although there is a little variable confusion in his/her answer. See below for my approach - largely following user51547's answer (any credits please give them to him/her) but amending it slightly.

Ingredients:

$a=e^{ix}\\ b=e^{iy}\\ c=e^{iz}\\ \omega=a+b+c=\frac{11}{5}+2i\\ \bar{{\omega}}=\bar{a}+\bar{b}+\bar{c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{11}{5}-2i\\ abc=e^{i(x+y+z)} = u+iv\\ \bar{{\omega}}^2 abc=\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}+2\omega \text{ , and also }\bar{{\omega}}^2 abc = (\frac{21}{25} -\frac{44}{5}i)(u+iv)\\ {\omega}^2 \cdot \frac{1}{abc}=\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2\bar{{\omega}}\\ $

$v$ to be determined.

$tan(x) + tan(y) + tan(z)=\frac{17}{6}$ translates into

$S=\frac{17i}{6}T$ where

$S=(a-\bar{a})(b+\bar{b})(c+\bar{c}) + (a+\bar{a})(b-\bar{b})(c+\bar{c}) + (a+\bar{a})(b+\bar{b})(c-\bar{c})\\ T=(a+\bar{a})(b+\bar{b})(c+\bar{c})$

Expanding and simplifying gives
$\frac{S}{i}=6Im(abc)+2Im(\bar{{\omega}}^2 abc) -4Im(\omega) = 6v+2(\frac{21}{25}v-\frac{44}{5}u)-8$ and
$T=2Re(abc)+2Re(\bar{{\omega}}^2 abc) -4Re(\omega)=2u+2(\frac{21}{25}u+\frac{44}{5}v)-\frac{44}{5}$

hence

$(36\cdot 25+6\cdot 42)v-88\cdot 6 \cdot 5u -8\cdot 6\cdot 25 =17[(50+42)u+88 \cdot 5v-44\cdot 5)]$

which gives

$1051u+1582v=635$

As $u^2+v^2=1$, the parametrization mentioned by user51547 which effectively uses $t=tan(\frac{x+y+z}{2})$

leads to

$(t-2)(843t+104)=0$

Hence $v=\frac{4}{5}$ or $v=\frac{2 \cdot -\frac{104}{843}}{1+(\frac{104}{843})^2)}=-\frac{175344}{721465}$