$\arctan(-3/2)$ doesn't give expected result.

Question

Let's say I want to find the angle measure (in degrees) such that $\tan(x) = -3/2$.

It turns out that $x \approx 123.7$, and when I compute $\tan(123.7)$, I get $\approx -3/2$; so far so good.

However, when I compute $\arctan(-3/2)$, I get $\approx -56.3$ where I expected $\approx 123.7$ since $\arctan$ is the inverse of $\tan$.

Answer 1

By definition, $\arctan x$ is the angle (in radians) between $-\pi/2$ and $\pi/2$ whose tangent is $x$.

There are infinitely many numbers $y$ such that $\tan y=-3/2$.

Answer 2

I get $−56.3$ where I expected $123.7$ since $\arctan$ is the inverse of $\tan$.

Hint: How much is $180^\circ-56.3^\circ$ ?

Answer 3

The points on the circle corresponding to $-56.3^\circ$ and to $123.7^\circ$ are antipodal to each other, since $-56.3^\circ+ 180^\circ= 123.7^\circ$. The period of the tangent and cotangent functions is only a half circle, i.e. $180^\circ$, rather than a full circle as with the sine, cosine, secant, and cosecant. Thus $\tan(-56.3^\circ)=\tan123.7^\circ$.

So the question is which of these two, $-56.3^\circ$ or $123.7^\circ$, should we take to be the inverse-tangent of $-3/2$? The picture below suggests that taking it to be the negative one allows the inverse tangent function to be continuous, since the tangent does not go up to infinity as the angle stays between $\pm90^\circ$.