Inverse Trigonometric Functions

In this section, we study the inverse of trigonometric function.

Arcsin $x$

Let $${[-\frac{\pi}{2},\frac{\pi}{2}]}$$ be the domain of the function $y = \sin{x}$. Then the function $y = \sin{x}$ becomes one-to-one. Thus we can think of the inverse function. We write this function as

$$y = \sin^{-1}{x} \$$   or$$\ y = \arcsin{x}$$

and call arcsin x.

NOTE The domain of $y = \sin^{-1}{x}$ is the range of $y = \sin{x}$. Thus we have $[-1,1]$. On the other hand, the range of $y = \sin^{-1}{x}$ is the domain of $y = \sin{x}$. Thus we have $${[-\frac{\pi}{2},\frac{\pi}{2}]}$$. In other words,

$$y = \sin^{-1}{x} \Leftrightarrow x = \sin{y} \ (-1 \leq x \leq 1, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2})$$

In this situation, the value in this range $[-\frac{\pi}{2},\frac{\pi}{2}]$ is called principal value of arcsin $x$.

Figure 1.4: arcsin x

Example 1..11   Find the value of $${\sin^{-1}(\frac{1}{2})}$$

SOLUTION Note that $${y = \sin^{-1}(\frac{1}{2})}$$ means $${\frac{1}{2} = \sin{y}}$$. Note also the values of $y$ must be in the interval $${[\frac{-\pi}{2}, \frac{\pi}{2}]}$$. Since $\sin{y}$ takes the value $\frac{1}{2}$ for $y = \frac{\pi}{6}\pm 2\pi,\frac{5\pi}{6} \pm 2\pi$, $y$ must be in $${[\frac{-\pi}{2}, \frac{\pi}{2}]}$$. Thus we have $${y = \frac{\pi}{6}}$$ $\ \blacksquare$

Exercise 1..11   Find the value of $${\sin^{-1}{(-\frac{1}{2})}}$$.

SOLUTION $y = \sin^{-1}{(-\frac{1}{2})}$ is equivalent to $\sin y = \frac{-1}{2}\ (-\frac{\pi}{2} \leq y \leq \frac{\pi}{2})$. Thus, $y = -\frac{\pi}{4}$ $\ \blacksquare$

Arccos$x$

Let the interval $[0,\pi]$ be the domain of $y = \cos{x}$. Then $y = \cos{x}$ becomes one-to-one. Thus we can think of the inverse function. We write this function as

$$y = \cos^{-1}{x} \ or \ y = \arccos{x}$$

and call this function arccos$x$.

NOTE The domain of the function $y = \cos^{-1}{x}$ is $[-1,1]$ which is the range of the function $y = \cos{x}$. The range of the function $y = \cos^{-1}{x}$ is $[0,\pi]$. Thus

$$y = \cos^{-1}{x} \Leftrightarrow x = \cos{y} \ (-1 \leq x \leq 1, 0 \leq y \leq \pi)$$

The value in the interval $[0,\pi]$ is called pv of arccos$x$.

Example 1..12   Find the value of $${\sin{\left(\cos^{-1}{\frac{1}{2}}\right)}}$$.

SOLUTION Since $x = \cos^{-1}{\frac{1}{2}}$ is equivalent to $\cos x = \frac{1}{2}$ for $0 \leq x \leq \pi$, we have $x = \frac{\pi}{3}$. Thus

$$\sin(\cos^{-1}\frac{1}{2}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\ensuremath{\ \blacksquare}$$

Exercise 1..12   Find the value $x$ which satisafies $\sin^{-1}\frac{1}{3} = \cos^{-1}{x}$.

SOLUTION Let $y = \sin^{-1}{\frac{1}{3}} = \cos^{-1}{x}$. Then

$$\frac{1}{3} = \sin{y}\ (-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}).$$

Since $y = \cos^{-1}{x}$, we have $x = \cos{y} \geq 0\ (0 \leq y \leq \pi)$. Note that,

$$x = \cos{y} = \sqrt{1 - \sin^{2}{y}} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}\ensuremath{\ \blacksquare}$$

ArcTan

Let the interval $${[-\frac{\pi}{2},\frac{\pi}{2}]}$$ be the domain of the function $y = \tan{x}$. Then the function $y = \tan{x}$ becomes one-to-one. Thus we can think of the inverse function. We write this function as

$$y = \tan^{-1}{x} \$$   or$$\ y = \arctan{x}$$

and say arctan$x$.

The domain of the function $y = tan^{-1}{x}$ is the range of the function $y = \tan{x}$ which is $(-\infty,\infty)$. The range is $${(-\frac{\pi}{2},\frac{\pi}{2})}$$. Thus we have

$$y = \tan^{-1}{x} \Leftrightarrow x = \tan{y} \ (-\infty < x < \infty, -\frac{\pi}{2} < y < \frac{\pi}{2})$$

The value in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ is called principal value of arctan$x$.

Example 1..13   Find the value of $x$ satisfies the following.
1. $\tan^{-1}{2} = \cos^{-1}{x}$2. $\sin^{-1}{\frac{1}{3}} = \cos^{-1}{x}$

SOLUTION 1. Set $y = \tan^{-1}{2} = \cos^{-1}{x}$. Then

$$2 = \tan{y}\ (-\frac{\pi}{2} < y < \frac{\pi}{2}).$$

Since $y = \cos^{-1}{x}$, we have $x = \cos{y} \geq 0\ (0 \leq y \leq \pi)$. Now dividing both sides of the identity $\cos^{2}{y} + \sin^{2}{y} = 1$ by $\cos^{2}{y}$ and noting $\tan{y} = \frac{\sin{y}}{\cos{y}}$, we have $1 + \tan^{2}{y} = \frac{1}{\cos^{2}{y}}$. Thus,

$$x = \cos{y} = \frac{1}{\sqrt{1 + \tan^{2}{y}}} = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}\ensuremath{\ \blacksquare}$$

2. Let $y = \sin^{-1}{\frac{1}{3}} = \cos^{-1}{x}$. Then

$$\frac{1}{3} = \sin{y}\ (-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}).$$

Since $y = \cos^{-1}{x}$, we have $x = \cos{y} \geq 0\ (0 \leq y \leq \pi)$. Substitute this into the identity $\cos^{2}{y} + \sin^{2}{y} = 1$, we have $x^2 + \sin^{2}{y} = 1$. Thus,

$$x = \sqrt{1 - \sin^{2}{y}} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}\ensuremath{\ \blacksquare}$$

Exercise 1..13   Show the identity $${\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}}$$ holds for all $x$.

SOLUTION Set $a = \sin^{-1}{x}, b = \cos^{-1}{x}$. Then we have

$$x = \sin a \ (-\frac{\pi}{2} \leq a \leq \frac{\pi}{2}), x = \cos b\ (0 \leq b \leq \pi).$$

Note that we can express $\cos b$ as $\cos b = \sin(\frac{\pi}{2} - b)$. Then $\sin a = \cos b = \sin(\frac{\pi}{2} - b)$. Since $-\frac{\pi}{2} \leq \frac{\pi}{2} - b \leq \frac{\pi}{2}$, the function $\sin{x}$ is one-to-one. Thus $a = \frac{\pi}{2} - b$. Since $a = \sin^{-1}{x}, b = \cos^{-1}{x}$, we obtain $\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}.$ $\ \blacksquare$

Exercise A

1.

Evaluate the followings?D

(a) $${\tan^{-1}{0}}$$ (b) $${\sin^{-1}{\frac{1}{2}}}$$ (c) $${\sin{\left(\cos^{-1}{\frac{1}{2}}\right)}}$$

Exercise B

1.

Find the value of the followings?D

(a) $${\sin^{-1}{(\frac{-1}{2})}}$$ (b) $${\cos^{-1}{(-1)}}$$ (c) $${\tan^{-1}{(-1)}}$$ (d) $${\tan^{-1}{\sqrt{3}}}$$

2.

Show $${\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}}$$ is true for all $x$?D

3.

Derive the following formulas?D

(a) $${\sin^{-1}{(-x)} = - \sin^{-1}{x}}$$ (b) $${\cos^{-1}{(-x)} = \pi - \cos^{-1}{x}}$$