Solving Systems of Trigonometric Equations

Systems of trigonometric equations can be solved algebraically by substituting each trigonometric function with a variable corresponding to its argument.

A Practical Example

Let’s work through the following system of trigonometric equations:

$$ \begin{cases} \sin x + \sin y = \frac{3}{2} \\ \\ \sin x - \sin y = - \frac{1}{2} \end{cases} $$

We substitute $ s = \sin x $ and $ t = \sin y $ to simplify the system:

$$ \begin{cases} s + t = \frac{3}{2} \\ \\ s - t = - \frac{1}{2} \end{cases} $$

The next step is to solve the system using a method of your choice, such as substitution.

From the first equation, we isolate $ s $:

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ s - t = - \frac{1}{2} \end{cases} $$

Now substitute $ s = \frac{3}{2} - t $ into the second equation:

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ \frac{3}{2} - t - t = - \frac{1}{2} \end{cases} $$

Simplify to find:

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ - 2t = - \frac{1}{2} - \frac{3}{2} \end{cases} $$

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ - 2t = - 2 \end{cases} $$

Eliminate the negative sign by multiplying through by $-1$:

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ 2t = 2 \end{cases} $$

Solve for $ t $:

$$ \begin{cases} s = \frac{3}{2} - t \\ \\ t = 1 \end{cases} $$

Substitute $ t = 1 $ back into the first equation to solve for $ s $:

$$ \begin{cases} s = \frac{3}{2} - 1 \\ \\ t = 1 \end{cases} $$

Simplify:

$$ \begin{cases} s = \frac{1}{2} \\ \\ t = 1 \end{cases} $$

The solutions to the algebraic system are:

$$ s = \frac{1}{2} $$

$$ t = 1 $$

Since $ s = \sin x $ and $ t = \sin y $:

$$ \sin x = \frac{1}{2} $$

$$ \sin y = 1 $$

Both values are within the range of the sine function, $[-1, 1]$.

The basic sine equations have solutions of the form $ \alpha + 2\pi k $ or $ \pi - \alpha + 2\pi k $:

$$ \alpha + 2\pi k \vee (\pi - \alpha) + 2\pi k $$

To find $ x $ and $ y $, we calculate the arcsine of both sides:

$$ \arcsin(\sin x) = \arcsin\left(\frac{1}{2}\right) $$

$$ \arcsin(\sin y) = \arcsin(1) $$

Taking the arcsine gives:

$$ x = \arcsin\left(\frac{1}{2}\right) $$

$$ y = \arcsin(1) $$

The arcsine of $ \frac{1}{2} $ is $ 30^\circ $ or $ \pi/6 $, while the arcsine of $ 1 $ is $ 90^\circ $ or $ \pi/2 $:

$$ x = \frac{\pi}{6} $$

$$ y = \frac{\pi}{2} $$

Thus, the solutions for $ x $, with $ \alpha = \pi/6 $, are:

$$ x = \alpha + 2\pi k \vee (\pi - \alpha) + 2\pi k $$

$$ x = \frac{\pi}{6} + 2\pi k \vee (\pi - \frac{\pi}{6}) + 2\pi k $$

$$ x = \frac{\pi}{6} + 2\pi k \vee \frac{5\pi}{6} + 2\pi k $$

For $ y $, with $ \alpha = \pi/2 $, the solutions are:

$$ y = \alpha + 2\pi k \vee (\pi - \alpha) + 2\pi k $$

$$ y = \frac{\pi}{2} + 2\pi k \vee (\pi - \frac{\pi}{2}) + 2\pi k $$

$$ y = \frac{\pi}{2} + 2\pi k $$

In conclusion, the solutions to the system are:

$$ x = \frac{\pi}{6} + 2\pi k \vee \frac{5\pi}{6} + 2\pi k \ \& \ y = \frac{\pi}{2} + 2\pi k $$

And so on.