How do I solve $\sin{a} + \sin{b} + \sin{c} = 2 \land \cos{a} + \cos{b} + \cos{c} = 2$?

How do I solve $$\sin{a} + \sin{b} + \sin{c} = 2 \land \cos{a} + \cos{b} + \cos{c} = 2$$?

I tried the following in Mathematica, but it did not give any solutions. I would appreciate an analytical solution (or the proof that no solutions exist).

Solve[{
Sin[a] + Sin[b] + Sin[c] == 2,
Cos[a] + Cos[b] + Cos[c] == 2},
{a, b, c}]

• I believe solutions (I have a harder time imaging uniqueness of it) exist, but I see no way of finding one analytically. Commented Jan 23, 2022 at 12:33
• Have you attempted to use "TrigReduce" prior to your "Solve" ? Commented Jan 23, 2022 at 14:06
• If you use the same syntax aasking to solve for $(a,b)$, a solution comes out and you have $a(c)$ and $b(c)$. If $a+b+c=\pi$, the problem is quite different. Commented Jan 27, 2022 at 8:45

What it looks to me is that you want to make $$(2,2)=\vec{a}+\vec{b}+\vec{c}$$ where $$\vec{a}, \vec{b}, \vec{c}$$ are unit vectors in the Euclidean plane. Here is a way to do it geometrically:

(a) Draw a circle around $$O=(0,0)$$ with radius $$1$$ and a circle around $$C=(2,2)$$ with radius $$2$$. They will intersect in two points $$X,Y$$. Take a point $$A$$ anywhere on the smaller arc $$\overset{\frown}{XY}$$, and let $$\vec{a}=\vec{OA}$$. $$\vec{a}$$ will be a unit vector.

(b) Now make two circles of radius $$1$$ centered in $$A$$ and $$C$$. They will intersect in one or two points: take any of them (call it $$B$$) and now both $$\vec{b}=\vec{AB}$$ and $$\vec{c}=\vec{BC}$$ will be unit vectors, and $$\vec{a}+\vec{b}+\vec{c}=\vec{OC}=(2,2)$$.

• Equivalently, we want three (complex) numbers on the unit circle summing to $2+2i$. Commented Jan 23, 2022 at 12:12
• Can you give a non-geometric solution? Commented Jan 23, 2022 at 13:23
• @HappyFace Once you have a solution, it is harder to think about another (radically different) one. Of course, this solution can be easily converted to an algebraic one, using tools from analytical geometry (equation of a circle). If it matters to you, give it a try. (Can you algebraically solve for the two points of intersection of two circles in general i.e. $(x-x_i)^2+(y-y_i)^2=r_i^2$ for $x_i,y_i,r_i$ given, $i=1,2$?) I will also be interested what the other answers to your question would look like.
– user700480
Commented Jan 23, 2022 at 14:59
• I think I have a solution. I will write it down tomorrow. Commented Jan 23, 2022 at 23:04

This question reminds me of articulated mechanisms (Watt, Peaucelier, Hart, Chebyshev...), for example the one that is represented at the top of this page of a huge compendium of such mechanisms.

It is why I still propose geometrical approaches (sorry...) that are efficient : they allow to obtain exact solutions (formulas (11)) or, using a different approach, approximate solutions (formulas (5)).

Let us write down for convenience the given equations:

$$\begin{cases}\cos a+\cos b+\cos c&=&2\\ \sin a+\sin b+\sin c&=&2\end{cases}\tag{1}$$

I would like to give a first geometrical representation of (1), with points

$$B=\binom{\cos a}{\sin a}, \ C=\binom{\cos a+\cos b}{\sin a+\sin b}, \ D=\binom{\cos a+\cos b+\cos c=2}{\sin a+\sin b+\sin c=2}\tag{2}$$

in a system of rectangular coordinates (Fig. 1).

One can see positions of $$A,B,C,D$$ in an ordinary case and (in red) $$A,B',C',D$$ representing a limit case.

Fig. 1.

Fig. 2 provides a representation of 3 surfaces, now in $$(a,b,c)$$ coordinate space:

$$\begin{cases}c&=&\cos^{-1}(2-\cos a - \cos b) & \ (blue) \\ c&=&\sin^{-1}(2-\sin a - \sin b) & \ (purple) \\ c&=&\pm \cos^{-1}(\tfrac{1}{4 \sqrt{2}}(7-2 \cos(a-b))) & \ (red)\end{cases}\tag{3}$$

Fig. 2.

The first and second equations come clearly from (1).

The third equation is issued from the squaring and adding of equations :

$$\begin{cases}2-\cos c&=&\cos a+\cos b\\ 2-\sin c&=&\sin a+\sin b\end{cases}\tag{4}$$

The interest of the third equation is that, apart from its cylindrical representation, it gives the following constraint:

$$\cos(a-b) \ge \frac12 (7-4 \sqrt{2})\approx 0.67157$$

that can be exploited later on.

Remark: In Fig. 2, we have taken a $$[0,\pi]^3$$ box in order to have a broader view, but in fact solutions to (1) belong to $$[0,\pi/2]^3$$. Had we taken an even larger scope, we would have seen that the surface represented by the second equation generates a bowl-like surface identical to the first surface, but for the fact that it is a bowl turned upside down.

Fig. 2 shows that these surfaces share a common curve ; otherwise said, the set of solutions $$(a,b,c)$$ is a one-dimensional loop, approximately circular, in the $$(a,b,c)$$ coordinate system.

Edit 1: on an experimental basis, I have found that the intersection curve (otherwise said, the general solution to equations (1)) is very close (see Fig. 3) to the 3D curve with equations:

$$\begin{cases}a&=&\dfrac{\pi}{4}+0.245 \cos t - 0.42 \sin t \\ b&=&\dfrac{\pi}{4}+0.245 \cos t + 0.42 \sin t \\ c&=&\dfrac{\pi}{4}-0.49\cos t\end{cases}\tag{5}$$

Otherwise said, an approximate solution of (1) (giving RHSides between $$1.99$$ and $$2.01$$ for any value of $$t$$) is:

$$\begin{pmatrix}a\\b\\c \end{pmatrix}=\begin{pmatrix}\dfrac{\pi}{4}\\ \dfrac{\pi}{4}\\ \dfrac{\pi}{4} \end{pmatrix}+\cos t \underbrace{\begin{pmatrix}+0.245\\+0.245\\-0.49 \end{pmatrix}}_U+\sin t \underbrace{\begin{pmatrix}-0.425 \\ +0.425 \\ 0 \end{pmatrix}}_V\tag{6}$$

With $$U,V$$ forming an orthogonal basis of plane $$(P)$$ with equation $$a+b+c=0$$; formulas (6) describe (up to numerical approximations) the circle with center $$(\pi/4,\pi/4,\pi/4)$$ and radius $$\approx 0.6$$ belonging to plane $$a+b+c=3 \pi/4$$.

There is a slight discrepancy with a pure circle: as we can see on Fig. 3, the resulting curve is partly in front of the two surfaces (the 3 continuous arcs) and partly behind them (the 3 dotted arcs).

Fig. 3.

Edit 2: Following exchanges with @dxiv, I have realized that a complex approach is rather direct.

The issue being to find triples $$(a,b,c)$$ such that

$$e^{ia}+e^{ib}+e^{ia}=2+2i=2\sqrt{2}e^{i \pi/4}\tag{7}$$

which, by setting

$$a=\pi/4+\alpha, \ b=\pi/4+\beta, \ c=\pi/4+\alpha\tag{8}$$

can be transformed into:

$$\underbrace{e^{i \alpha}+e^{i \beta}}_{2z}+e^{i \gamma}=2\sqrt{2}\tag{9}$$

where $$z$$ is the midpoint of the line segment joining $$e^{i \alpha}$$ and $$e^{i \beta}$$ (represented as point $$D$$ on Fig. 4) ; as this line segment is orthogonal to the line joining $$O$$ and $$z$$, we can write, using (9) and what we just said above:

$$\begin{cases}z&:=&\frac23 \sqrt{2}-\tfrac12 e^{i \gamma}=\cos \phi e^{i \delta}& \ (point D)\\ e^{i \alpha}&:=&e^{i \delta}e^{i \phi}& \ (point A)\\ e^{i \beta}&:=&e^{i \delta}e^{-i \phi}& \ (point B)\end{cases}\tag{10}$$

Fig. 4: The blue point is the barycenter $$(\frac23 \sqrt{2}, \ 0)$$ of points $$A=e^{i \alpha}, \ B=e^{i \beta}, \ C=e^{i \gamma}, \$$. The materialized angle has value $$\phi$$.

Being given angle $$\gamma$$, using relationships (8), we are able to deduce the two other angles $$\alpha$$ and $$\beta$$:

$$\alpha,\beta=\tan^{-1}\left(\frac{\sin \gamma}{\cos \gamma -2 \sqrt{2} }\right)\pm \cos^{-1}\left(\sqrt{\frac94-\sqrt{2} \cos \gamma}\right)\tag{11}$$

Fig. 5. The curves of $$\alpha,\beta$$ (in green and red) as functions $$\alpha(\gamma),\beta(\gamma)$$ of variable $$\gamma$$ given by (11). One notices the perfect symmetry between $$a$$ and $$b$$, with an almost elliptic curve.

(and then adding $$\pi/4$$ to retrieve values of $$a,b,c$$ (formulas (8)).

Rather simple calculations show that angles $$\alpha, \beta, \gamma$$ should be in the range $$[-L,L]$$ where $$L=\cos^{-1}(\tfrac58 \sqrt{2})=27.89°$$.

• Nicely done from a novel perspective (+1). The parametric $a,b$ at the end could also be written as $\,0.49 \cos(t \pm \varphi)\,$ for some $\,\varphi\,$, and I am pretty sure the errors are just numerical roundoffs.
– dxiv
Commented Jan 26, 2022 at 3:28
• Impressive and beautiful ! Commented Jan 27, 2022 at 8:49

The following will consider the slightly more general form:

\begin{align} \cos{a} + \cos{b} + \cos{c} &= u \tag{1} \\ \sin{a} + \sin{b} + \sin{c} &= v \tag{2} \end{align}

Let $$\,\sigma = u + i v\,$$, $$\,\alpha=\cos a + i \sin a\,$$ and similar for $$\,\beta,\gamma\,$$ with $$\,|\alpha|=|\beta|=|\gamma|=1\,$$, then adding $$\,(1) + i\,\cdot (2)\,$$ translates to $$\,\alpha+\beta+\gamma= \sigma\,$$. Taking conjugates and using that $$\,\overline\alpha = 1 / \alpha\,$$ etc allows solving for $$\,\alpha\,$$, $$\,\beta\,$$ in terms of $$\,\sigma\,$$, $$\,\gamma\,$$:

\begin{align} \alpha + \beta &= \sigma - \gamma \\ \frac{1}{\alpha}+\frac{1}{\beta} &= \overline\sigma-\overline\gamma \;\;\implies\;\; \alpha\beta = \frac{\alpha+\beta}{\overline\sigma-\overline\gamma} = \frac{\sigma - \gamma}{\overline\sigma-\overline\gamma} \end{align}

With $$\displaystyle\,\lambda=\frac{1}{\sigma-\gamma}\,$$ it follows that $$\,\alpha\,$$, $$\,\beta\,$$ are the roots of the quadratic:

\begin{align} t^2 - \frac{1}{\lambda} t + \frac{\overline\lambda}{\lambda} = 0 \;\;\;\;\iff\;\;\;\; &\lambda t^2 - t+ \overline \lambda = 0 \\ \;\;\;\;\iff\;\;\;\; & \alpha,\beta = \frac{1 \pm i\sqrt{4 |\lambda|^2-1}}{2\lambda} \tag{3} \end{align}

The roots $$\,\alpha,\beta\,$$ will have modulus $$\,1\,$$ iff $$\,4 |\lambda|^2 \ge 1\,$$ $$\,\iff |\sigma-\gamma| \le 2\,$$. Therefore the system has solutions when at least one $$\,\gamma\,$$ exists such that $$\,|\gamma|=1\,$$ and $$\,|\sigma-\gamma| \le 2\,$$, which is equivalent to $$\,|\sigma| \le 1+2=3\,$$. In that case, for each such $$\,\gamma\,$$ there exists a solution $$\,\alpha\,$$, $$\,\beta\,$$ given by $$\,(3)\,$$, which is unique up to a permutation of variables.

[ EDIT ] $$\;$$ Inspired by Jean Marie's answer, here is a different way to parameterize the solution set.

The condition $$\,|\sigma-\gamma| \le 2\,$$ was previously established. Moreover, the triangle inequality gives the additional constraints $$\,\big||\sigma|-1\big|\le|\sigma-\gamma|\le |\sigma|+1\,$$. Then $$\,\sigma-\gamma\,$$ can be written in the form:

$$\sigma-\gamma = 2\omega\,\cos\varphi \; \begin{cases} |\omega|=1 \\ \varphi \in [\varphi_{min}, \varphi_{max}] \subseteq [0, \pi/2] \; \begin{cases} \varphi_{min}= \arccos \min\left(\frac{|\sigma|+1}{2}, 1\right) \\ \varphi_{max}=\arccos \frac{\big||\sigma|-1\big|}{2} \tag{4} \end{cases} \end{cases}$$

For each $$\,\varphi \in [\varphi_{min}, \varphi_{max}]\,$$ there will exist two $$\,\omega\,$$ such that $$\,|\omega|=1\,$$ and $$\,|\sigma - 2 \omega \cos \varphi|$$ $$= |\gamma| = 1\,$$, which are the roots of the equation below (it can be shown that the quadratic always has two complex roots $$\,\omega_{1,2}\,$$ of modulus $$\,|\omega_1|=|\omega_2|=1\,$$ when $$\,\varphi\,$$ is in the given range):

$$|\sigma - 2 \omega \cos \varphi|^2=1 \iff 2 \overline \sigma \cos \varphi \cdot \omega^2 - \left(|\sigma|^2 + 4 \cos^2 \varphi - 1\right)\cdot\omega + 2\sigma\cos\varphi = 0 \tag{5}$$

With either $$\,\sigma-\gamma=2\omega \cos\varphi\,$$, $$\,\omega \in \{\omega_1, \omega_2\}\,$$ it follows that $$\,2\lambda = \frac{1}{\omega\cos\varphi}\,$$, then $$\,(3)\,$$ becomes:

$$\alpha,\beta = \frac{1 \pm i\sqrt{\frac{1}{\cos^2\varphi}-1}}{\frac{1}{\omega\cos\varphi}} = \omega\left(\cos\varphi \pm i \sin\varphi\right)$$

So, in the end, the solution set is:

\begin{cases} \begin{align} \alpha &= \omega(\cos\varphi + i \sin\varphi) \\ \beta &= \omega(\cos\varphi - i \sin\varphi) \\ \gamma &= \sigma - 2\omega\cos\varphi \qquad\qquad\qquad \text{where}\;\; \varphi \in [\varphi_{min}, \varphi_{max}] \text{ from (4)}\,, \;\,\omega \text{ from (5)} \end{align} \end{cases}

• [+1] It has taken me some time before understanding your (rather short) solution "drawn" (in almost mechanical terms) by $\gamma$ (i.e., angle $c$) It should be linked to my "solution" (see my Edit). Commented Jan 25, 2022 at 23:59
• @JeanMarie Thanks, though I am not entirely happy with the end result. It is technically correct and it does solve the problem, but it loses (or, at least, obfuscates) the inherent symmetry in it. The other approach was trying to preserve that, but it did not work out in the end. After seeing your solution I'll have to take another look at maybe parameterizing $\,\gamma\,$ in a more meaningful way, which is indeed what drives everything else in my algebraic approach.
– dxiv
Commented Jan 26, 2022 at 3:36
• P.S. $\,$ The other approach was about the cubic with roots $\,\alpha\,$, $\,\beta\,$, $\,\gamma\,$ which turns out to be $\,p(z) = z^3 - \sigma z^2 + \tau \overline\sigma z - \tau\,$ where $\,\tau = \alpha\beta\gamma\,$ is constrained by the condition that $\,p(z)\,$ has all three roots on the unit circle. The condition is equivalent to $\,p^{\,\prime}(z)\,$ having both roots on or inside the unit circle [1] but I was not able to carry that step through to completion. If anyone sees a way to do it, please @ ping me so that I can +1 that answer.
– dxiv
Commented Jan 26, 2022 at 6:21
• @JeanMarie Not the same, but the edit gets now to something closer to your form. It also has a direct geometric interpretation - start with two unit vectors $\,\alpha,\beta\,$ at angle $\,2\varphi\,$, then find the rotation $\,\omega\,$ that brings their sum to distance $\,1\,$ from $\,\sigma\,$.
– dxiv
Commented Jan 26, 2022 at 6:44
• @JeanMarie Yes, that closes the circle ;-)
– dxiv
Commented Jan 28, 2022 at 3:37