Inverse Trigonometry: Finding Principal Values

This lesson explains how to find the principal value of an inverse trigonometric function. It starts from the definitions and the domain and range table, then works through examples with positive and negative inputs, and finishes with composite expressions like $\sin^{-1}(\sin x)$.

Definitions of the inverse functions

If $\sin x = y$ then $x = \sin^{-1} y$, and the same pattern gives the other five inverse functions: $\cos^{-1} y$, $\tan^{-1} y$, $\csc^{-1} y$, $\sec^{-1} y$, and $\cot^{-1} y$. Note that $\sin^{-1} y$ means the inverse function, not the reciprocal $\tfrac{1}{\sin y}$.

Domain and range (principal value branches)

The range of each inverse function is its principal value branch.

$\sin^{-1} x:\quad [-1, 1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$

$\cos^{-1} x:\quad [-1, 1] \to [0, \pi]$

$\tan^{-1} x:\quad \mathbb{R} \to \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$

$\cot^{-1} x:\quad \mathbb{R} \to (0, \pi)$

$\csc^{-1} x:\quad \mathbb{R} \setminus (-1, 1) \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \setminus \{0\}$

$\sec^{-1} x:\quad \mathbb{R} \setminus (-1, 1) \to [0, \pi] \setminus \left\{\tfrac{\pi}{2}\right\}$

Principal values for positive inputs

Example. $\sin^{-1} \tfrac{1}{\sqrt{2}}$. Let $y = \sin^{-1}\tfrac{1}{\sqrt{2}}$, so $\sin y = \tfrac{1}{\sqrt{2}} = \sin\tfrac{\pi}{4}$, giving $y = \tfrac{\pi}{4}$. Since $\tfrac{\pi}{4} \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, the principal value is $\tfrac{\pi}{4}$.

Example. $\cos^{-1} \tfrac{\sqrt{3}}{2}$. Here $\cos y = \tfrac{\sqrt{3}}{2} = \cos\tfrac{\pi}{6}$, so $y = \tfrac{\pi}{6}$, which lies in $[0, \pi]$. The principal value is $\tfrac{\pi}{6}$.

Example. $\csc^{-1} \sqrt{2}$. Here $\csc y = \sqrt{2} = \csc\tfrac{\pi}{4}$, so $y = \tfrac{\pi}{4}$, which lies in the range. The principal value is $\tfrac{\pi}{4}$.

Signs of the ratios in each quadrant

To handle negative inputs, recall where each ratio is positive:

• First quadrant: all ratios positive.
• Second quadrant: $\sin$ and $\csc$ positive.
• Third quadrant: $\tan$ and $\cot$ positive.
• Fourth quadrant: $\cos$ and $\sec$ positive.

For a reference angle $\theta$, the angle in the second quadrant is $\pi - \theta$, in the third is $-\pi + \theta$, and in the fourth is $-\theta$. Choose the quadrant whose angle lands inside the function's range.

Principal values for negative inputs

Example. $\sin^{-1}\left(-\tfrac{1}{2}\right)$. Here $\sin y = -\tfrac{1}{2}$, with reference angle $\theta = \tfrac{\pi}{6}$ since $\sin\tfrac{\pi}{6} = \tfrac{1}{2}$. Sine is negative in the third and fourth quadrants. The third quadrant gives $-\pi + \tfrac{\pi}{6} = -\tfrac{5\pi}{6}$, which is outside $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$. The fourth quadrant gives $-\theta = -\tfrac{\pi}{6}$, which is inside. So the principal value is $-\tfrac{\pi}{6}$.

Example. $\cot^{-1}\left(-\sqrt{3}\right)$. Here $\cot y = -\sqrt{3}$, reference angle $\theta = \tfrac{\pi}{6}$. Cotangent is negative in the second and fourth quadrants. The second quadrant gives $\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$, which lies in $(0, \pi)$. The principal value is $\tfrac{5\pi}{6}$.

Example. $\tan^{-1}(-1)$. Here $\tan y = -1$, reference angle $\theta = \tfrac{\pi}{4}$. Tangent is negative in the second and fourth quadrants. The second quadrant gives $\tfrac{3\pi}{4}$ (outside the range); the fourth gives $-\tfrac{\pi}{4}$, which lies in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. The principal value is $-\tfrac{\pi}{4}$.

Example. $\sec^{-1}\left(-\tfrac{2}{\sqrt{3}}\right)$. Here $\sec y = -\tfrac{2}{\sqrt{3}}$, reference angle $\theta = \tfrac{\pi}{6}$. Secant is negative in the second and third quadrants. The second quadrant gives $\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$, which lies in the range $[0, \pi] \setminus \left\{\tfrac{\pi}{2}\right\}$. The principal value is $\tfrac{5\pi}{6}$.

Inverse of a trig function of an angle

While $\sin^{-1}(\sin x) = x$ only when $x$ lies in the principal range, otherwise rewrite the inner angle first.

Example. $\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$. Since $\tfrac{2\pi}{3}$ is outside $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, use $\sin\left(\pi - \theta\right) = \sin\theta$ to write $\sin\tfrac{2\pi}{3} = \sin\tfrac{\pi}{3}$. Then $\sin^{-1}\left(\sin\tfrac{\pi}{3}\right) = \tfrac{\pi}{3}$, which is in range.

Example. $\sin^{-1}\left(\sin\tfrac{3\pi}{5}\right)$. Here $\tfrac{3\pi}{5} = 108^\circ$ is outside the range, so $\sin\tfrac{3\pi}{5} = \sin\left(\pi - \tfrac{3\pi}{5}\right) = \sin\tfrac{2\pi}{5}$. Since $\tfrac{2\pi}{5} = 72^\circ$ is in range, the answer is $\tfrac{2\pi}{5}$.

Example. $\tan^{-1}\left(\tan\tfrac{7\pi}{6}\right)$. Since $\tfrac{7\pi}{6}$ is outside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ and $\tan\left(\pi + \theta\right) = \tan\theta$, we have $\tan\tfrac{7\pi}{6} = \tan\tfrac{\pi}{6}$, so the answer is $\tfrac{\pi}{6}$.

Key takeaways

• The principal value is the angle whose ratio matches the input and that lies inside the function's range.
• For negative inputs, find the reference angle, then pick the quadrant whose angle falls inside the range, using $\pi - \theta$, $-\pi + \theta$, or $-\theta$.
• For $\sin^{-1}(\sin x)$ and similar, first rewrite the inner angle so it sits in the principal range before simplifying.