Inverse Trigonometric Functions

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

Arcsin
Let $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$ 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 $\displaystyle y = \sin^{-1}{x} \$ or $\displaystyle \ y = \arcsin{x}$ and call arcsin x.

Arcsin

Let $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$ 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

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

and call arcsin x.

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

$\displaystyle y = \sin^{-1}{x} \Leftrightarrow x = \sin{y} \ (-1 \leq x \leq 1, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2})$
In this situation, the value in this range $[-\frac{\pi}{2},\frac{\pi}{2}]$ is called principal value of arcsin .

Figure 1.4: arcsin x

$\includegraphics[width=8cm]{SOFTFIG-1/Fig1-2-3.eps}$

Example 1..11 Find the value of $\displaystyle{\sin^{-1}(\frac{1}{2})}$
SOLUTION Note that $\displaystyle{y = \sin^{-1}(\frac{1}{2})}$ means $\displaystyle{\frac{1}{2} = \sin{y}}$ . Note also the values of must be in the interval $\displaystyle{[\frac{-\pi}{2}, \frac{\pi}{2}]}$ . Since $\sin{y}$ takes the value $\frac{1}{2}$ for $y = \frac{\pi}{6}\pm 2\pi,\frac{5\pi}{6} \pm 2\pi$ , must be in $\displaystyle{[\frac{-\pi}{2}, \frac{\pi}{2}]}$ . Thus we have $\displaystyle{y = \frac{\pi}{6}}$ $\ \blacksquare$

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

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
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

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

$\displaystyle y = \cos^{-1}{x} \Leftrightarrow x = \cos{y} \ (-1 \leq x \leq 1, 0 \leq y \leq \pi)$
The value in the interval $[0,\pi]$ is called pv of arccos.

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

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

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

$\displaystyle \frac{1}{3} = \sin{y}\ (-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}).$
Since $y = \cos^{-1}{x}$ , we have $x = \cos{y} \geq 0\ (0 \leq y \leq \pi)$ . Note that,

$\displaystyle x = \cos{y} = \sqrt{1 - \sin^{2}{y}} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}\ensuremath{\ \blacksquare}$
ArcTan
Let the interval $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$ 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

$\displaystyle y = \tan^{-1}{x} \$ or $\displaystyle \ y = \arctan{x}$
and say arctan.
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 $\displaystyle{(-\frac{\pi}{2},\frac{\pi}{2})}$ . Thus we have

$\displaystyle y = \tan^{-1}{x} \Leftrightarrow x = \tan{y} \ (-\infty < x < \infty, -\frac{\pi}{2} < y < \frac{\pi}{2})$
The value in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ is called principal value of arctan.

Example 1..13 Find the value of 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

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

$\displaystyle x = \cos{y} = \frac{1}{\sqrt{1 + \tan^{2}{y}}} = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}\ensuremath{\ \blacksquare}$
2. Let $y = \sin^{-1}{\frac{1}{3}} = \cos^{-1}{x}$ . Then

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

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

Exercise 1..13 Show the identity $\displaystyle{\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}}$ holds for all .
SOLUTION Set $a = \sin^{-1}{x}, b = \cos^{-1}{x}$ . Then we have

$\displaystyle x = \sin a \ (-\frac{\pi}{2} \leq a \leq \frac{\pi}{2}), x = \cos b\ (0 \leq b \leq \pi).$
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) $\displaystyle{\tan^{-1}{0}}$ (b) $\displaystyle{\sin^{-1}{\frac{1}{2}}}$ (c) $\displaystyle{\sin{\left(\cos^{-1}{\frac{1}{2}}\right)}}$

Exercise B

1.

Find the value of the followings?D
(a) $\displaystyle{\sin^{-1}{(\frac{-1}{2})}}$ (b) $\displaystyle{\cos^{-1}{(-1)}}$ (c) $\displaystyle{\tan^{-1}{(-1)}}$ (d) $\displaystyle{\tan^{-1}{\sqrt{3}}}$

2.

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

3.

Derive the following formulas?D
(a) $\displaystyle{\sin^{-1}{(-x)} = - \sin^{-1}{x}}$ (b) $\displaystyle{\cos^{-1}{(-x)} = \pi - \cos^{-1}{x}}$

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