A calculus of the absurd

7.4 General solutions to trigonometric equations

When is this equation true?

\begin{equation*} \sin (x) = \frac {\sqrt {3}}{2} \hspace {5pt} 0 < x < 2\pi \end{equation*}

Using the spangles (in the previous section), we know that it’s true when \(x=\frac {\pi }{3}\). In the interval in question, however, this isn’t the only point where it’s true! If you look at the graph of \(\sin (x)\), it’s clear that there’s another solution to this in the interval \(\left (\frac {\pi }{2}, \pi \right )\). Because of \(\sin \)’s symmetry, we know that this solution will be at \(\pi - \frac {\pi }{3}\).

We can generalise this to any point (and for different trig functions). For any point \(a\) which is in the range of \(\sin (x)\), we know

\begin{equation} \sin (\theta ) = a \implies \theta = \arcsin (a) \end{equation}

We can call the value returned by any inverse trig function the “principle value”. It is usually the angle closest to \(0\). However, this value is not necessarily the only possible value. For one, we know that \(\sin (x)\) and \(\cos (x)\) repeat every \(2\pi \) (or \(360^{\circ }\)) so for any value of \(\theta \) which solves \(\sin (\theta ) = a\) or \(\cos (\theta ) = a\), then that value plus or minus any multiple of \(2\pi \) (or \(360^{\circ }\)) will also solve the equation.

The other thing to bear in mind is that \(\sin (\theta )\) has an axis of symmetry in the lines \(x=\frac {\pi }{2}\) and \(x=\frac {3\pi }{2}\). This means that if \(\theta \) is the principal value solving \(\sin (\theta )=a\), then \(\sin (\pi - \theta ) = a\) is also a solution.

For \(\cos (\theta )\), something very similar is the case, except that the axis of symmetry is in the line \(x=\pi \), and thus if \(\theta \) is the principal value solving \(\cos (\theta )=a\), then \(\cos (2\pi - \theta ) = a\) is also a solution.

Overall, we can write that for \(\sin (\theta ) = a\)

\begin{align} \theta = \begin{cases} \begin{aligned} & \arcsin (a) \pm 2\pi \cdot k \\ & \pi - \arcsin (a) \pm 2\pi \cdot k \end {aligned} & k \in \mathbb {N} \end {cases} \end{align}

And that for \(\cos (\theta )\)

\begin{align} \theta = \begin{cases} \begin{aligned} & \arccos (a) \pm 2\pi \cdot k \\ & 2\pi - \arccos (a) \pm 2\pi \cdot k \end {aligned} & k \in \mathbb {N} \end {cases} \end{align}

This is also equivalent to

\begin{align} \theta = \pm \arccos (a) \pm 2\pi \cdot k \hspace {12pt} k \in \mathbb {N} \end{align}