Inverse functions reverse the action of their counterparts. For a function $f$ that maps $x$ to $y$, its inverse, $f^{-1}$, maps $y$ back to $x$. For trigonometric functions, this concept requires careful handling. Since functions like $\sin(x)$ are periodic, they are not one-to-one over their entire domain. An output like $0.5$ corresponds to infinitely many inputs (e.g., $\pi/6$, $5\pi/6$, $13\pi/6$, etc.).

To define a valid inverse, we must restrict the domain of the original trigonometric function to a principal branch where it is one-to-one. This ensures that each output value corresponds to a unique input angle. The resulting inverse functions are often called arcfunctions, such as $\arcsin(x)$ for the inverse of $\sin(x)$. Graphically, an inverse function is a reflection of the original function across the line $y=x$.

The three primary inverse trigonometric functions are arcsine, arccosine, and arctangent.

Note the different ranges. The range for $\arcsin(x)$ covers quadrants I and IV, yielding angles where the cosine is positive. The range for $\arccos(x)$ covers quadrants I and II, providing angles where the sine is positive. For $\arctan(x)$, the range is identical to that of $\arcsin(x)$ but excludes the endpoints, as $\tan(x)$ is undefined at $\pm\pi/2$.

The inverses of the reciprocal trigonometric functions—cosecant, secant, and cotangent—are defined similarly by restricting their domains.

A key property is their relationship to the primary inverse functions. For example, $\arccsc(x) = \arcsin(1/x)$ and $\arcsec(x) = \arccos(1/x)$ for $x$ in their respective domains. The identity for arccotangent is slightly more nuanced: $\arccot(x) = \arctan(1/x)$ for $x > 0$, but $\arccot(x) = \arctan(1/x) + \pi$ for $x < 0$. This adjustment ensures the output lies within the correct principal range of $(0, \pi)$.

Inverse trigonometric functions are essential for solving trigonometric equations. When an equation is in the form $\sin(x) = c$, the principal solution is $x = \arcsin(c)$. To find all solutions, we must account for the periodic nature of the sine function.

Consider solving the equation $2\sin^2(\theta) - 5\sin(\theta) + 2 = 0$. This is a quadratic in terms of $\sin(\theta)$. Factoring gives $(\sin\theta - 2)(2\sin\theta - 1) = 0$.

The solution $\sin\theta = 2$ is impossible, since the range of sine is $[-1, 1]$.

The other solution is $\sin\theta = 1/2$. The principal value is $\theta = \arcsin(0.5) = \pi/6$. The general solution is $\theta = n\pi + (-1)^n (\pi/6)$.

In physics, inverse functions appear in problems involving oscillations, waves, and projectile motion. For instance, the range $R$ of a projectile launched with initial velocity $v_0$ at an angle $\theta$ is given by $R = \frac{v_0^2 \sin(2\theta)}{g}$. To find the angle required to achieve a specific range, we rearrange the formula:

In electrical engineering, the phase angle $\phi$ in an RLC circuit is given by $\tan(\phi) = \frac{X_L - X_C}{R}$, where $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance, and $R$ is the resistance. The phase angle, which describes the offset between voltage and current, is found using the arctangent: $\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$.

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Understanding the precise definitions and restricted ranges of inverse trigonometric functions is crucial for their correct application in both theoretical and practical contexts.