Solving 3 equations involving trigonometric expressions.

Question

I have a question which was asked in IMO,

Solve the system of equations:

$x \sin a+y\sin 2a+z\sin 3a=\sin 4a$

$x \sin b+y\sin 2b+z\sin 3b=\sin 4b$.

$x \sin c+y\sin 2c+z\sin 3c=\sin 4c$.

I am clueless at it, any hint/help is appriciated.

Thanks

Answer 1

We assume that the initial system has a non-zero determinant (see remark below for a discussion).

Let us consider the following system:

$$\tag{1}\left\{ \begin{aligned} x e^{ia} + y e^{2ia} + ze^{3ia}&=e^{4ia}\\ x e^{ib} + y e^{2ib} + ze^{3ib}&=e^{4ib}\\ x e^{ic} + y e^{2ic} + ze^{3ic}&=e^{4ic} \end{aligned} \right. \qquad\iff\qquad \begin{cases} \begin{aligned} x + y e^{ia} + z e^{2ia} &= e^{3ia}\\ x + y e^{ib} + z e^{2ib} &= e^{3ib}\\ x + y e^{ic} + z e^{2ic} &= e^{3ic} \end{aligned}\end{cases} $$

Let us define polynomial $P(T):=x+yT+zT^2-T^3$.

With this notation, system $(1)$ is equivalent to

$$P(e^{ia})=P(e^{ib})=P(e^{ic})=0.$$

The three roots of $P$ being $e^{ia}$, $e^{ib}$, $e^{ic}$, Vieta's formulas give:

$$\tag{2} \begin{cases} \begin{aligned} x&= \phantom{-}e^{ia}e^{ib}e^{ic}=e^{i(a+b+c)}\\ y&= -(e^{i(a+b)}+e^{i(a+c)}+e^{i(b+c)})\\ z&= \phantom{-}e^{ia}+e^{ib}+e^{ic} \end{aligned} \end{cases} $$

It suffices now to take the imaginary part of system (1) whose solutions will be the imaginary parts of solutions above, i.e.,

$$\tag{3} \begin{cases} \begin{aligned} x&= \phantom{-}\sin(a+b+c)\\ y&=-\bigl(\sin(a+b)+\sin(a+c)+\sin(b+c)\bigr)\\ z&= \phantom{-}\sin(a)+\sin(b)+\sin(c) \end{aligned} \end{cases} $$

Remark: the determinant of system $(1)$, under its second form, is a Vandermonde determinant; it is non-zero unless, two values among $e^{ia}$, $e^{ib}$, $e^{ic}$ are identical, which means two angles $a,b,c$ are equal modulo $2\pi$. This is the only case of indeterminacy.

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Edit: a different proof, not using complex numbers, although using the same 3rd degree polynomial trick is as follows.

Consider the following formulas:

$$\begin{cases}\sin 2a&=&\sin a \times (2 \alpha)\\ \sin 3a&=&\sin a \times (4 \alpha^2-1)\\ \sin 4a&=&\sin a \times (8 \alpha^3-4\alpha)\end{cases} \ \ \ \text{with} \ \ \ \alpha:=\cos a$$

which all are particular cases of the fact that $\frac{\sin(na)}{\sin(a)}$ can be expressed as a degree $n-1$ polynomial in $\cos(a)$ (which is named 2nd order Chebyshev polynomial).

Plugging these values of $\sin 2a$, $\sin 3a$,$\sin 4a$ in the first equation of the initial system, and simplifying by $\sin a$ (assumed non-zero) we obtain

$$x+y2\alpha+z(4 \alpha^2-1)=8 \alpha^3-4\alpha.$$

In other words,

$$8 \alpha^3-4z\alpha^2-(2y+4)\alpha+(z-x)=0$$

We can do the same for the second and third equations, with $\beta:=\cos b$ and $\gamma:=\cos c$. Gathering these 3 results, we have:

$$\tag{4}\left\{ \begin{aligned} 8 \alpha^3-4z\alpha^2-(2y+4)\alpha+(z-x)=0\\ 8 \beta^3-4z\beta^2-(2y+4)\beta+(z-x)=0\\ 8 \gamma^3-4z\gamma^2-(2y+4)\gamma+(z-x)=0 \end{aligned} \right.$$

This means that $\alpha,\beta,\gamma$ are roots of polynomial:

$$P(T):=8T^3-4zT^2-(2y+4)T+(z-x)$$

Viete's formulas give:

$$\tag{5} \begin{cases} \begin{aligned} z/2&=&\alpha+\beta+\gamma\\ (2y+4)/8&=&\alpha\beta+\beta\gamma+\gamma\alpha\\ (z-x)/8&=&\alpha\beta\gamma \end{aligned} \end{cases} \ \ \ \implies$$

$$\begin{cases} \begin{aligned} x&=& 2(\cos(a)+\cos(b)+\cos(c))+8\cos(a)\cos(b)\cos(c)\\ y&=&-2-4(\cos(a)\cos(b)+\cos(b)\cos(c)+\cos(c)\cos(a))\\ z&=& 2(\cos(a)+\cos(b)+\cos(c))\\ \end{aligned} \end{cases} $$

Remark: it must be understood that this is the same set of solutions as presented in the yellow cartridge above, expressed in terms of cosines instead of sines of angles $a,b,c$.

Answer 2

Assuming you're trying to solve for $(x, y, z)^{T}$ then

Hint: Rewrite $$ \begin{pmatrix} \sin a & \sin 2a & \sin 3a\\ \sin b & \sin 2b & \sin 3b\\ \sin c & \sin 2c & \sin 3c\\ \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} \sin 4a \\\sin 4b \\ \sin 4c \end{pmatrix}$$ Thus $$ \begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} \sin a & \sin 2a & \sin 3a\\ \sin b & \sin 2b & \sin 3b\\ \sin c & \sin 2c & \sin 3c\\ \end{pmatrix}^{-1}\begin{pmatrix} \sin 4a \\\sin 4b \\ \sin 4c \end{pmatrix}$$ The next step is to compute the inverse, using double angle formula for $\sin$.