We can see from the graph that \(y=\tan x\) has a period of \(\pi\). This means that \(\tan(x+n\pi) = \tan x\) for all integers \(n\).

Therefore \(\tan A=\tan B=a\) if and only if \(A = B + n\pi\) for some integer \(n\). In other words, \(A\) and \(B\) can only differ by an integer multiple of \(\pi\).

If, for example, \(\tan A =\tan B = 1\), then \(A = \dfrac{\pi}{4}+n\pi\) for some integer \(n\). We’d also have \(B = \dfrac{\pi}{4}+m\pi\) for some integer \(m\), and so \[A = B + k\pi\quad \text{where $k$ is an integer.}\]

More generally, if \(\tan A=\tan B = a\), then \(A = \arctan a + n\pi\) and \(B = \arctan a + m\pi\), where \(n\) and \(m\) are integers, so \(A = B+ k\pi\) where \(k\) is an integer.

Please note that \(\arctan\) and \(\tan^{-1}\) mean the same thing and both are widely used in books and online. Similarly, you may see \(\arcsin\) or \(\sin^{-1}\) and \(\arccos\) or \(\cos^{-1}\).

If \(\cos A=\cos B\), what can we say about \(A\) and \(B\)?

Here is a graph of \(y=\cos x\), showing the solutions of \(\cos x=a\).

We can see from the graph that \(y=\cos x\) has a period of \(2\pi\). This means that \(\cos(x+2n\pi) = \cos x\) for all integers \(n\).

Looking at the graph above, solutions to \(\cos x=a\) fall into two types (what we might call the ‘left-hand solutions’ and the ‘right-hand solutions’). For example, if \(a = \dfrac{1}{2}\), then we can call \(\dfrac{\pi}{3}\) a right-hand solution, and \(-\dfrac{\pi}{3}\) a left-hand solution.

This means that any solution to \(\cos x = a\) is of the form \[\text{(a right-hand solution)}+ 2n\pi \quad \text{or} \quad \text{(a left-hand solution)} + 2n\pi \quad \text{for some integer $n$.}\]

For example, if \(\cos x= \dfrac{1}{2}\), then \(x = \dfrac{\pi}{3} + 2n\pi\) or \(x = -\dfrac{\pi}{3} + 2n\pi\) for some integer \(n\).

Therefore, if \(\cos A = \cos B = a\), then \[A = \arccos a + 2n\pi \quad \text{or} \quad A = -\arccos a + 2n\pi,\] and \[B = \arccos a + 2m\pi \quad \text{or} \quad B = -\arccos a + 2m\pi,\]

where \(m\) and \(n\) are integers. This means that \(A = B + 2k\pi\) or \(A = -B + 2k\pi\), where \(k\) is an integer.

If \(\sin A=\sin B\), what can we say about \(A\) and \(B\)?

Here is a graph of \(y=\sin x\), showing the solutions of \(\sin x=a\).

The period of \(\sin x\) is also \(2\pi\) and again the solutions to \(\sin x=a\) fall into two types. For example, if \(a = \tfrac{1}{2}\), then we can call \(\tfrac{\pi}{6}\) a left-hand solution and \(\pi -\tfrac{\pi}{6}\) a right-hand solution. Any solution to \(\sin x = a\) is therefore of the form \[\text{(a right-hand solution)}+ 2n\pi \quad \text{or} \quad \text{(a left-hand solution)} + 2n\pi \quad\text{for some integer $n$.}\]

If \(\sin A = \sin B = a\), then we can say \(A\) and \(B\) have the following forms \[A = \arcsin a + 2n\pi \quad \text{or} \quad A = \pi -\arcsin a + 2n\pi,\quad \text{where $n$ is an integer,}\]\[B = \arcsin a + 2m\pi \quad \text{or} \quad B = \pi -\arcsin a + 2m\pi \quad \text{where $m$ is an integer.}\]

This means \(A = B + 2k\pi\) or \(A = \pi-B + 2k\pi\), where \(k\) is an integer.

We can think of \(y=\sin x\) as a translation of \(y=\cos x\), by \(\tfrac{\pi}{2}\) in the positive \(x\) direction. The symmetry of \(y=\cos x\) about the \(y\)-axis then means \(y=\sin x\) is symmetric about \(x = \tfrac{\pi}{2}\), and the solutions for \(\sin x = a\) are symmetrical about this line. Therefore we could write the solutions of \(\sin x=a\) as \[x = \left(\arcsin a - \dfrac{\pi}{2}\right) + 2n\pi + \dfrac{\pi}{2} \quad \text{or} \quad x = \left(\dfrac{\pi}{2} - \arcsin a\right) + 2n\pi + \dfrac{\pi}{2}.\]