# Solving a trigonometric system of equations

I am a math tutor at the local university and encountered a trigonometric problem which stumped me. (Of course, I'm certain that this material was covered in my students class, but they still didn't know how to solve it, either.) Basically, the problem boils down to solving the following system of equations for $x$ and $y$ with $-\pi \leq x, y \leq \pi$:

$c_1\cos x + c_2 \cos y = 0$

$c_1\sin x + c_2 \sin y = 0$

How to solve a system of trigonometric equations is a similar question, but I am having difficulty following the answer and applying them to this particular problem. Any tips would be appreciated.

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Are $c_1,c_2$ the same between the two equations? –  Ross Millikan Oct 14 '13 at 16:32
@RossMillikan Yes –  Code-Guru Oct 14 '13 at 16:39

You could just multiply the first equation by $\sin{x}$ and the second by $\cos{x}$ and subtract to get

$$c_2 \sin{x} \cos{y} - c_2 \cos{x} \sin{y} = c_2 \sin{(x-y)} = 0$$

Then $x=y$ or $x-y = k \pi$, $k \in \mathbb{Z}$. Plug this back into either equation to determine $x$ or $y$.

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Case $1:$ If $c_1,c_2\ne0$ are same, between the two equations

we have $c_1\cos x=-c_2\cos y\ \ \ \ (1)$ and $c_1\sin x=-c_2\sin y\ \ \ \ (1=2)$

Divide to get $\tan x=\tan y\implies x=n\pi+y$ where $n$ is any integer

Put the value of $x$ in $(1)$ or $(2)$

We need to deal the even and the odd values of $n$ separately

Case $2$: If $c_1,c_2\ne0$ are not same, between the two equations

HINT:

Get the values of $\sin x,\cos x$ from $(1),(2)$

Square & add to eliminate $x$ and form an equation in $y$ only and solve

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Eliminate $c_1$ from the two equations, and you get $c_2$ times some trig equals zero. Do you remember the formulas for $\sin(x\pm y)$?

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I think OP is solving for $x$ and $y$ –  Ross Millikan Oct 14 '13 at 16:34
@RossMillikan That is correct. I should have specified that. –  Code-Guru Oct 14 '13 at 16:38
"Do you remember the formulas for sin(x±y)?" Vaguely. I need to look them up for a refresher. –  Code-Guru Oct 14 '13 at 16:41

Writing the equations as \begin{align} c_1 \cos x &= - c_2 \cos y &(1)\\ c_1 \sin x &= - c_2 \sin y &(2) \end{align} we square and add to get \begin{align} c_1^2 \left( \cos^2 x + \sin^2 x \right ) &= \phantom{\pm} c_2^2 \left( \cos^2 y + \sin^2 y \right) \\ \implies \qquad c_1^2 &= \phantom{\pm}c_2^2 \\ \implies \qquad c_1 &= \pm c_2 \end{align} Since $c_1 = c_2 = 0$ makes the whole system disappear, we assume the constants are non-zero, and we divide the system's equations by $c_1$ to get ... \begin{align} \cos x &= \pm \cos y &(1^\prime)\\ \sin x &= \pm \sin y &(2^\prime) \end{align} where the "$\pm$"s match. So, either $$\cos x = \cos y \qquad \text{and} \qquad \sin x = \sin y \qquad (a)$$ or $$\cos x =-\cos y \qquad \text{and} \qquad \sin x =- \sin y \qquad (b)$$

• $(a)$ simply implies that $x=y$ (given your restricted domain).
• $(b)$ implies that the points $(\cos x, \sin x)$ and $(\cos y, \sin y)$ are diametrically opposed on the unit circle, so that $y = x \pm \pi$ (with the "$\pm$" chosen as necessary to keep $y$ in your chosen domain).
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