We now discuss how to solve equations of the form, say,

$LaTeX: \tan \theta = 1.3$

for \(\theta\). In particular, this will be useful when considering the polar form of complex numbers.

We begin with an example which is constructed so that we can use the table of special values for the trigonometric functions from Module 1.3.

Example 6.19

Let us solve

$LaTeX: \tan \theta = \sqrt{3}$

Geometrically, this can be interpreted as asking us to find all angles \(\theta\) so that the line from the origin, having this angle, has slope equal to \(\sqrt{3}\).

Now, since \(\sqrt{3}\) appears in the table of special values for the tangent (see Module 1.3), we know that one solution is \(\theta = \pi/3\). By considering the unit circle, we know that by adding or subtracting any integer multiple of \(\pi\) from the angle, its slope remains the same. Therefore, all angles

$LaTeX: \theta = \frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z}$

solve the equation \(\tan \theta = \sqrt{3}\).

Fig. 4: Both the angles \(\pi/3\) and \(\pi/3 + \pi\) solve \(\tan \theta = \sqrt{3}\).

Here is a similar example for the sine function.

Example 6.20

Let us now solve the equation

$LaTeX: \sin x = \sqrt{3}/2.$

By the table of values for the trigonometric functions in Module 1.3, we know that \(x=\pi/3\) is one solution of this equation. Moreover, by the figure below, we see that \(x = \pi - \pi/3 = 2\pi/3\) is a second solution. Since these are the only solutions for \(x\) in the interval \([0,2\pi)\), it follows by the periodicity of the sine, that all solutions are given by

$LaTeX: x = \pi/3 + 2\pi n$

and

$LaTeX: x = 2\pi/3 + 2\pi n$

where \(n\) is any integer.

Fig. 5. The angles \(x=\pi/3\) and \(x= \pi-\pi/3\) give the same value for the sine.

But the question we ask here is really the following: how to solve examples, such as the ones above, if we cannot use the table of special values?

Here is how to do this for the sine function.

Example 6.21

To solve the equation

$LaTeX: \sin \theta = 0.9$

most high school textbooks would ask you to use the \(\sin^{-1}\) or \(\arcsin\) function in your calculator (warning: here, \(\sin^{-1}\) is not the same as \(1/\sin\), which is why we will always write \(arcsin\) to avoid confusion). This function can take any number from \([-1,1]\) as input, and will give the unique angle in the 1st or 2nd quadrant that solves \(\sin \theta = 0.9\). In this case, the answer (in radians) is

$LaTeX: \theta = \arcsin(0.9) = 1.11977...$

To understand what happens, consider the following figure.

Fig. 6: The angle shown solves \(\sin \theta = 0.9\) since the blue line hits the unit circle at a point with x-coordinate equal to \(0.9\). Notice the same is true for the dotted line, which has angle \(\pi-\theta\).

From the figure, we also see that if \(\theta = 1.11977...\) solves the equation, then by symmetry, so does \(\pi-\theta = 2.02182...\). Moreover, adding or subtracting integer multiples of \(2\pi\) changes nothing, so all solutions are given by

$LaTeX: \theta = \left\{ \begin{aligned} 1.11977... + 2\pi k, \quad k \in \mathbb{Z} \\[2mm] 2.02182... + 2\pi k, \quad k \in \mathbb{Z} \end{aligned}\right.$

Next, we consider the cosine.

Example 6.22

We now solve the equation

$LaTeX: \cos \theta = -0.3.$

This time, we use the \(\cos^{-1}\) or \(\arccos\) button in your calculator. Similar to \(\arcsin\), this function can take any number in \([-1,1]\) as input, but now gives the unique angle in the 1st or 4th quadrants that solves \(\cos \theta = -0.3\). In this case, the answer that pops out (again, in radians) is

$LaTeX: \theta = \arccos(-0.3) = 1.875489...$

To understand what happens, consider the following figure.

Fig. 7: The angle shown solves \(\cos \theta = -0.3\) since the blue line hits the unit circle at a point with y-coordinate equal to \(-0.3\). Notice the same is true for the dotted line, which has angle \(-\theta\).

From the figure, we also see that if \(\theta = 1.875489..\) solves the equation, then by symmetry, so does \(- \theta = -1.875489...\). Moreover, adding or subtracting integer multiples of \(2\pi\) changes nothing, so all solutions are given by

$LaTeX: \theta = \pm 1.875489... + 2\pi k, \quad k \in \mathbb{Z}.$

Finally, we consider the situation with the tangent. However, to be able to explain what is going on, we first explain a pleasant visual way of interpreting the tangent function for angles in the 1st or 4th quadrants.

Remark 6.23 (Visual interpretation of the tangent function)

Recall that in terms of the unit circle, \(\tan \theta\) was defined to be the slope of the line segment from the origin having the angle \(\theta\). If we suppose that \(\theta\) is in the 1st or 4th quadrant, then if we follow this line until it has \(x\)-coordinate equal to \(1\), then the \(y\)-coordinate will be exactly equal to the slope.

Vid. 5: The line with angle \(\theta\) intersects the vertical line at \(x=-1\) at the height \(y = \tan \theta\).

With the above remark in place, we are now ready to consider our final example.

Example 6.24

We now solve the equation

$LaTeX: \tan \theta = 1.3.$

As before, we use the \(\tan^{-1}\) or \(\arctan\) button on our calculator. This function can take any point on the real line as input, and gives the unique angle in the 1st or 4th quadrant that solves \(\tan \theta = 1.3\), namely:

$LaTeX: \theta = \arctan(1.3) = 0.915101...$

To understand what happens, we use the above geometric interpretation of the tangent.

Fig. 8: The angle shown solves \(\tan \theta = 1.3\) since the blue line hits the red line at \(x=1\) at a point with y-coordinate equal to \(1.3\). Notice the same is true for the dotted line, which has angle \(\theta + \pi\).

Finally, as in the first example above, if \(\theta = 0.915101...\) solves the equation, then all solutions are given by

$LaTeX: \theta = 0.915101... + \pi k, \quad k \in \mathbb{Z}.$

Finally, we mention that one of the major themes of the course MATA21 is to understand exactly how these values of the inverse trigonometric functions \(\arcsin, \arccos\) and \(\arctan\) can be computed. In particular, this will be one of the fruits of the beautiful theory of Taylor expansions.