Inverse Trigonometric Functions

Inverse trigonometric functions, like any other inverse function, are mathematical operators that undo the function's operation. For the right triangle

we have seen the basic trigonometric functions

$$\begin{array}{lcr} \sin \theta = \frac{b}{c}, & \qquad \cos \theta = \frac{a}{c}, & \qquad \tan \theta = \frac{b}{a}, \end{array}$$

where the angle is the input (or argument) to the function and the ratio of sides is the output of the function result. The inverse function reverses this operation, taking the ratio of sides as input and returning the angle as output:

$$\begin{aligned} \sin^{-1} \left( \frac{b}{c} \right) &= \theta \\ \cos^{-1} \left( \frac{a}{c} \right) &= \theta \\ \tan^{-1} \left( \frac{b}{a} \right) &= \theta. \end{aligned}$$

This means the inverse trigonometric functions are useful whenever we know the sides of a triangle and want to find its angles.

Note: The notation $\sin^{-1}$ might be confusing, as we normally use a negative exponent to indicate the reciprocal. However, in this case, $\sin^{-1} \alpha \neq \frac{1}{\sin \alpha}$. When we want the reciprocal of $\sin$ we use $\csc$ (see the wiki Reciprocal Trigonometric Functions for more details). In order to avoid this ambiguity, sometimes people might choose to write the inverse functions with an arc prefix. For example,

$$\arcsin \beta = \sin^{-1} \beta.$$

Notice that by definition of the inverse function, we have the following relationships:

$$\begin{aligned} \sin\big( \sin^{-1}(x)\big) & = x\\ \cos\big( \cos^{-1}(x)\big) & = x\\ \tan\big( \tan^{-1}(x)\big) & = x. \end{aligned}$$

If we restrict $x$ to be in the appropriate domain, the above function composition order can also be reversed:

$$\begin{aligned} \sin^{-1}\big( \sin(x)\big) & = x \qquad \text{ for } - \frac{\pi}{2} \leq x < \frac{\pi}{2}\\ \cos^{-1}\big(( \cos(x)\big) & = x \qquad \text{ for } 0 \leq x < \pi\\ \tan^{-1}\big(( \tan(x)\big) & = x \qquad \text{ for } - \frac{\pi}{2} \leq x < \frac{\pi}{2}. \end{aligned}$$

In the sine function, many different angles $\theta$ map to the same value of $\sin(\theta)$. For example,

$$0 = \sin 0 = \sin (\pi) = \sin (2\pi) = \cdots = \sin (k \pi)$$

for any integer $k$. To overcome the problem of having multiple values map to the same angle for the inverse sine function, we will restrict our domain before finding the inverse. The domain and ranges of the basic inverse trigonometric functions are as follows:

$$\begin{array}{|c|c|c|} \hline \text{Function} & \text{Domain} & \text{Range}\\ \hline \sin^{-1}(x) & [-1, 1] & \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]\\ \hline \cos^{-1}(x) & [-1, 1] & \left[ 0, \pi \right]\\ \hline \tan^{-1}(x) & (-\infty, \infty) & \left(- \frac{\pi}{2}, \frac{\pi}{2} \right)\\ \hline \end{array}$$

The graphs of the inverse functions are the original function in the domain specified above, which has been flipped about the line $y=x$. The effect of flipping the graph about the line $y=x$ is to swap the roles of $x$ and $y$, so this observation is true for the graph of any inverse function. See the wiki Inverse Trigonometric Graph for more details.

There are specific values of the inverse function that are useful to remember. By remembering the specific values of trigonometric functions, these values for the inverse function are also easily remembered:

$$\begin{array} {| c | c | c | c | c | c |} \hline \text{x} & \frac {\sqrt{0}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{4}} {2} \\ \hline \sin^{-1} (x) & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} \\ \hline \cos^{-1}(x) & \frac{\pi}{2} & \frac{\pi}{3} & \frac{\pi}{4} & \frac{\pi}{6} & 0\\ \hline \end{array}$$

Simplify the expression

$$\sin \left(\cos^{-1} \Big( \frac{x}{x+1} \Big) \right).$$

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Let $\theta = \cos^{-1} \left( \frac{x}{x+1} \right)$. Then by definition, $\cos(\theta) = \frac{x}{x+1}$. From the definition of cosine based on right triangles, construct the right triangle

Since $\cos \theta = \frac{a}{c},$ by letting $a=x$ and $c=x+1$, we have $\cos(\theta) = \frac{x}{x+1}$. From the Pythagorean theorem, we then have

$$b^2 = c^2 - a^2 = (x+1)^2-x^2 = x^2 + 2x + 1 - x^2 = 2x + 1.$$

This implies

$$\sin \left( \cos^{-1} \frac{x}{x+1} \right) = \sin (\theta) = \frac{b}{c} = \frac{\sqrt{2x + 1}}{x+1}.\ _\square$$

In the following equation, where $a$ and $b$ are coprime positive integers, what is the sum of $a$ and $b?$

$$\cos^{-1} \frac{17}{\sqrt{1130}}= \tan^{-1} \frac{a}{b}$$

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Since we are trying to find $a$ and $b$, we should take the tangent of both sides of the equation:

$$\begin{aligned} \tan \left( \cos^{-1} \frac{17}{\sqrt{1130}} \right) &= \tan \left( \tan^{-1} \frac{a}{b} \right) \\ \tan \Bigg( \underbrace{\cos^{-1} \frac{17}{\sqrt{1130}}}_{\large{\theta}} \Bigg) &= \frac{a}{b}. \end{aligned}$$

Further, we can use the ratio given to sketch the triangle with $\theta$ in it, using the definition of $\cos^{-1}$:

Right triangle with side 17 and hypotenuse square root of 1130

Now, using the Pythagorean theorem, we can see that $17^2 + a^2 = 1130$, which implies $a=\sqrt{1130-289}=29$. Finally, we evaluate $\tan \theta = \frac{29}{17}$, which implies $a+b=46$. $_\square$