Inverse Trigonometric Functions

Principal value ranges

Since $\sin$, $\cos$, etc. repeat forever, their inverses are only defined by restricting to one principal value range. For $\cot^{-1}x$, that range is the open interval $(0,\pi)$: note it's open, unlike $\tan^{-1}$ which uses $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

Worked example: evaluating a mixed expression

Find the value of $\sin\left(\dfrac{\pi}{2} + \cos^{-1}\dfrac{1}{\sqrt5}\right)$.

Solution: Let $\theta = \cos^{-1}\dfrac{1}{\sqrt5}$, so $\cos\theta = \dfrac{1}{\sqrt5}$. Using $\sin\left(\tfrac{\pi}{2}+\theta\right) = \cos\theta$:

Worked example: a principal value

Find the principal value of $\cot^{-1}\left(-\dfrac{1}{\sqrt3}\right)$.

Solution: Look for an angle in $(0,\pi)$ whose cotangent is $-\tfrac{1}{\sqrt3}$. Since $\cot\tfrac{\pi}{3} = \tfrac{1}{\sqrt3}$ and cotangent is negative in the second quadrant:

Worked example: a domain fact

What is the domain of $y = \sec^{-1}x$?

Solution: Secant only takes values with $|\sec\theta| \ge 1$, so its inverse can only accept inputs with $|x|\ge1$. The domain is $\mathbb{R} - (-1,1)$: every real number except the open interval strictly between $-1$ and $1$.

Worked example: proving a complementary identity

Prove that $\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$ for every real $x$.

Solution: Let $\theta = \tan^{-1}x$, so $\tan\theta = x$. Using the co-function identity $\tan\theta = \cot\left(\tfrac{\pi}{2}-\theta\right)$:

Substituting back $\theta = \tan^{-1}x$ gives exactly $\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$.