3. Trigonometry

e. Inverse Trig Functions

3. Inverse Tangent & Cotangent

Arc Tangent

The tangent function is not one-to-one, since for example \[ \tan(\theta+\pi)=\tan(\theta) \] We pick the branch with \(-\,\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\).

def_arctan1

The inverse function of \(\tan\) is \(\arctan\) (read “arc tangent”) or \(\tan^{-1}\) (read “inverse tangent”) which satisfies \[ \arctan(z)=\theta \qquad \text{where} \qquad z=\tan(\theta) \] provided   \(-\,\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\).

Notice that \(-\,\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\) is quadrants IV and I only.

def_arctan

Compute each of the following.

  1. \(\arctan\dfrac{1}{\sqrt{3}}\)

    \(\arctan\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)

  2. \(\arctan(-\sqrt{3})\)

    \(\arctan(-\sqrt{3})=-\,\dfrac{\pi}{3}\)

  3. \(\arctan(-1)\)

    \(\arctan(-1)=-\,\dfrac{\pi}{4}\)

Arc Cotangent

The cotangent function is not one-to-one, since for example \[ \cot(\theta+\pi)=\cot(\theta) \] We pick the branch with \(0 \lt \theta \lt \pi\).

def_arccot1

The inverse function of \(\cot\) is \(\text{arccot}\) (read “arc cotangent”) or \(\cot^{-1}\) (read “inverse cotangent”) which satisfies \[ \text{arccot}(z)=\theta \qquad \text{where} \qquad z=\cot(\theta) \] provided   \(0 \lt \theta \lt \pi\).

Notice that \(0 \le \theta \le \pi\) is quadrants I and II only.

def_arccot

Compute each of the following.

  1. \(\text{arccot}\,\dfrac{1}{\sqrt{3}}\)

    \(\text{arccot}\,\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{3}\)

  2. \(\text{arccot}(-\sqrt{3})\)

    \(\text{arccot}(-\sqrt{3})=\dfrac{5\pi}{6}\)

    If you answered \(-\,\dfrac{\pi}{6}\), you are correct that \(\cot\dfrac{-\pi}{6}=-\sqrt{3}\), but you have the wrong quadrant. \(\text{arccot}(z)\) always returns a value in the \(1^\text{st}\) or \(2^\text{nd}\) quadrant. \(-\,\dfrac{\pi}{6}\) is in the \(4^\text{th}\) quadrant.

  3. \(\text{arccot}(-1)\)

    \(\text{arccot}(-1)=\dfrac{3\pi}{4}\)

    If you answered \(-\,\dfrac{\pi}{4}\), you are correct that \(\cot\dfrac{-\pi}{4}=-1\), but you have the wrong quadrant. \(\text{arccot}(z)\) always returns a value in the \(1^\text{st}\) or \(2^\text{nd}\) quadrant. \(-\,\dfrac{\pi}{6}\) is in the \(4^\text{th}\) quadrant.

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