Question

Show that \(f\) is one-to-one on the indicated interval and therefore has an inverse function on that interval. $$ f(x)=\cot x \quad (0, \pi) $$

Short answer

This proof of contradiction has shown that \(f(x)=\cot x\) is one-to-one on the interval (0,π). As such, it has an inverse in this interval, which is \(f^{-1}(x) = \cot^{-1} x\).

Step-by-step solution

Understanding the Function

In this step, an understanding of the cotangent function, \(f(x) = \cot x\), is necessary. The cotangent function is the reciprocal of the tangent function and is defined as \(\cot x = \frac{\cos x}{\sin x}\). This function cycles between negative infinity and positive infinity in the interval (0,π).

Assume \(f\) is Not One-to-One

Assume, for the sake of contradiction, that \(f\) is not one-to-one. This means there exist \(x_1, x_2\) in (0,π) such that \(x_1 ≠ x_2\) and \(f(x_1) = f(x_2)\). That is, \(\cot x_1 = \cot x_2\). Since \(\cot x = \frac{\cos x}{\sin x}\), the assumption implies that \(\frac{\cos x_1}{\sin x_1} = \frac{\cos x_2}{\sin x_2}\). Setting these two equal to one another yields the equation: \(\cos x_1 \sin x_2 = \cos x_2 \sin x_1\)

Use Trigonometric Identity

Invoke the identity \(\sin (a - b) = \sin a \cos b - \cos a \sin b\). Plugging \(x_1\) for \(a\) and \(x_2\) for \(b\) gives \(\sin(x_1 - x_2) = \sin x_1 \cos x_2 - \cos x_1 \sin x_2\). Keep in mind that the sin function cannot yield zero in the interval (0, π).

Proof of Contradiction

In step 3, \(\sin x_1 \cos x_2 - \cos x_1 \sin x_2 = 0\) which can also be rewritten as-\(\sin(x_1 - x_2) = 0\), this only occurs when \(x_1 - x_2\) is a multiple of π. But this contradicts our assumption that \(x_1, x_2\) are different and lie in the interval (0,π), meaning \(f\) is one-to-one.

Finding the Inverse Function

Because \(f\) has been shown to be one-to-one on the interval (0, π), it will have an inverse function on that interval. The inverse of cotangent is arc cotangent, \(f^{-1}(x) = \cot^{-1} x\).