inverse trigonometric functions

The inverse trigonometric functions are \arcsin(x), \arccos(x) and \arctan(x). These functions perform the reverse operations to the original trigonometric functions \sin(x), \cos(x) and \tan(x) respectively. Recall that a function is invertible if it is one-to-one. Click here to revise inverse functions. Hence, before we can sketch the graphs of the inverse trigonometric functions, we must choose a domain for them for which they are one-to-one. Note that the original trigonometric functions work on angles and so each of the inverse trigonometric functions will return an angle. We use radians for all angles in the following – see more on radians. Also note that we use \arcsin(x) instead of \sin^{-1}(x), for example, as this can be confused with \text{cosec}(x), the reciprocal trigonometric function

Inverse Trigonometric Function: arcsin(x)

Since \sin(x) is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be \frac{-\pi}{2}\leq x\leq \frac{\pi}{2}. The range of \sin(x) is -1\leq\sin(x)\leq 1.  It follows that the domain of \text{arcsin}(x) is -1\leq x\leq 1 and the range is \frac{-\pi}{2}\leq \text{arcsin}(x)\leq \frac{\pi}{2}. The graphs of \sin(x) and \text{arcsin}(x) are reflections of one another in the line y=x.

Inverse Trigonometric Function: arccos(x)

Since \cos(x) is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be 0\leq x\leq \pi. The range of \cos(x) is -1\leq\cos(x)\leq 1.  It follows that the domain of \text{arccos}(x) is -1\leq x\leq 1 and the range is 0\leq \text{arcsin}(x)\leq \pi. The graphs of \cos(x) and \text{arccos}(x) are reflections of one another in the line y=x.

Inverse Trigonometric Function: arctan(x)

Since \tan(x) is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be \frac{-\pi}{2}< x<\frac{\pi}{2}. The range of \tan(x) is all of {\mathbb R}.  It follows that the domain of \text{arctan}(x) is x\in{\mathbb R} and the range is \frac{-\pi}{2}< \text{arctan}(x)< \frac{\pi}{2}. The graphs of \tan(x) and \text{arctan}(x) are reflections of one another in the line y=x.

Examples

Find the values (in radians) of the following without using a calculator:

  1. \text{arcsin}(1)
  2. \text{arccos}(\cos(\pi/7))
  3. \sin(\text{arctan}(1))

Solution:

  1. We can see from the graph or use the fact that \sin\left(\frac{\pi}{2}\right)=1 on this interval to find \text{arcsin}(1)=\frac{\pi}{2}.
  2. \text{arccos} and \cos are inverses of one another and so the result is \pi/7.
  3. \text{arctan}(1)=\frac{\pi}{4} and \sin(\pi/4)=\frac{\sqrt{2}}{2}. Hence \sin(\text{arctan}(1))=\frac{\sqrt{2}}{2}.

Sketch the graph of y=2\text{arccos}(-x), stating its domain and range.

Solution:

Recall from the transformations page that replacing x with -x reflects the original graph across the y-axis. Multiplying by 2 stretches the graph by a factor of 2 in the y-direction. It follows that the graph of y=2\text{arccos}(-x) is given by:

inverse trigonometric functions

The domain of y=2\text{arccos}(-x) is -1\leq x\leq 1 and the range is 0\leq y\leq 2\pi.

Given that \theta=\text{arccos}(x) where \theta is an acute angle, find an expression for \sin(\theta) in terms of x.

Solution:

If \theta=\text{arccos}(x) then \cos(\theta)=x. Using \sin^2(\theta)+\cos^2(\theta)=1 (see more trigonometric identities), \sqrt{1-\sin^2(\theta)}=x and so 1-\sin^2(\theta)=x^2. It follows that \sin(\theta)=\pm\sqrt{1-x^2}. Note that x=\cos(\theta) is between -1 and 1 (for acute angles it is actually between 0 and 1) and so 0\leq x^2\leq 1. It follows that in the expression for \sin(\theta) we are square rooting a nonnegative number, as we would hope. However, for acute \theta, \sin(\theta) must be positive and so \sin(\theta)=\sqrt{1-x^2}.