Domain, Range & Principal Value

Inverse trigonometric functions are used to find the angle when the value of a trigonometric ratio is known. Since trigonometric functions are not one-one over their full domains, their domains are restricted so that inverse functions can be defined properly. This is why domain, range, and principal value are central ideas in this chapter.

The inverse trigonometric functions are the inverse forms of trigonometric functions after suitable domain restriction. They are written as:

• \[\sin^{-1} x\]

• \[\cos^{-1} x\]

• \[\tan^{-1} x\]

• \[\cot^{-1} x\]

• \[\sec^{-1} x\]

• \[\csc^{-1} x\]

Important:

• \[\sin^{-1} x\]

does not mean 1/sin⁡x. It means the angle whose sine is x.

The value returned by an inverse trigonometric function is called its principal value. It is the unique angle chosen from the standard restricted interval for that function.

• Inverse trigonometric functions give angles corresponding to known trigonometric values.

• Their domains are restricted because ordinary trigonometric functions are not one-one on full domains.

• Principal value means the standard angle selected from a fixed interval.

Find Principal Values:

\[\mathrm{(i)}\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\]

Let \[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = x \]

For \[\cos^{-1} x\], principal value range is: \[ 0 \leq x \leq \pi \]

Now,

\[ \cos x = -\frac{1}{\sqrt{2}} \]

We know: \[ \cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}} \]

Hence,

\[ \boxed{\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}} \]

(ii) \[\tan^{-1}(-\sqrt{3})\]

Let \[ \tan^{-1}(-\sqrt{3}) = x \]

For \[\tan^{-1} x\], principal value range is: \[ -\frac{\pi}{2} < x < \frac{\pi}{2} \]

Now,

\[ \tan x = -\sqrt{3} \]

We know: \[ \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \]

Hence,

\[ \boxed{\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}} \]

(iii) \[\cot^{-1}(-1)\]

Let

\[ \cot^{-1}(-1) = x \] For \[\cot^{-1} x\], principal value range is: \[ 0 < x < \pi \]

Now,

\[ \cot x = -1 \]

We know: \[ \cot \frac{3\pi}{4} = -1 \]

Hence,

\[ \boxed{\cot^{-1}(-1) = \frac{3\pi}{4}} \]

Find the domain:

(i) \[\cos^{-1}(x^2 - 4)\]

1. Condition: The input of \[\cos^{-1}(u)\] must fall within \[[-1, 1]\], so:

\[ -1 \leq x^2 - 4 \leq 1 \]

2. Isolate \[x^2\]: Add 4 to all parts of the inequality:

\[ 3 \leq x^2 \leq 5 \]

3. Solve for \[x\]: Taking the square root yields \[\sqrt{3} \leq |x| \leq \sqrt{5}\], which splits into two intervals:

1. For positive values: \[x \in [\sqrt{3}, \sqrt{5}]\]
2. For negative values: \[x \in [-\sqrt{5}, -\sqrt{3}]\]

4. Final Domain: Combining both intervals gives:

\[ x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}] \]

(ii) \[\sec^{-1}(3x - 1)\]

1. Condition: The input \[u\] for \[\sec^{-1}(u)\] must satisfy \[|u| \geq 1\] (meaning \[u \leq -1\] or \[u \geq 1\]):

\[ 3x - 1 \leq -1 \quad \text{or} \quad 3x - 1 \geq 1 \]

2. Solve the first inequality:

\[ 3x \leq 0 \implies x \leq 0 \]

3. Solve the second inequality:

\[ 3x \geq 2 \implies x \geq \frac{2}{3} \]

4. Domain:

\[ x \in (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right) \]

Find the domain and range of \[f(x) = \cos^{-1}(1 - x^2)\]

Part 1: Finding the Domain

For \[\cos^{-1}(1 - x^2)\] to be defined:

\[ -1 \leq 1 - x^2 \leq 1 \]

Subtracting 1 from all sides:

\[ -2 \leq -x^2 \leq 0 \]

Multiplying by \[-1\] (reverses the inequality signs):

\[ 0 \leq x^2 \leq 2 \]

Taking the square root:

\[ -\sqrt{2} \leq x \leq \sqrt{2} \]

Therefore, Domain = \[[-\sqrt{2}, \sqrt{2}]\]

Part 2: Finding the Range

Since \[x \in [-\sqrt{2}, \sqrt{2}]\]:

\[ 0 \leq x^2 \leq 2 \]

Transforming it into \[1 - x^2\]:

\[ -1 \leq 1 - x^2 \leq 1 \]

This covers the entire input interval \[[-1, 1]\] for \[\cos^{-1}\]. Since \[\cos^{-1}([-1,1]) = [0, \pi]\]:

Therefore, Range = \[[0, \pi]\]