`f(x)=sin|x|+sin^(-1)(tanx)+sin(sin^(-1)x)`

Answer

To find the domain of the function \( f(x) = \sin |x| + \sin^{-1}(\tan x) + \sin(\sin^{-1} x) \), we will analyze each component of the function separately.

### Step 1: Determine the domain of \( f_1(x) = \sin |x| \)

The sine function is defined for all real numbers. Therefore, since \( |x| \) is also defined for all \( x \), we conclude that:

\[ \text{Domain of } f_1(x) = \mathbb{R} \quad \text{(all real numbers)} \]

**Hint:** The sine function is continuous and defined for all real values.

### Step 2: Determine the domain of \( f_2(x) = \sin^{-1}(\tan x) \)

The function \( \sin^{-1}(a) \) is defined only when \( a \) is in the interval \([-1, 1]\). Thus, we need to find when:

\[ -1 \leq \tan x \leq 1 \]

The tangent function is periodic with a period of \( \pi \). The values of \( x \) for which \( \tan x = -1 \) and \( \tan x = 1 \) occur at:

- \( \tan x = -1 \) at \( x = n\pi - \frac{\pi}{4} \) for integers \( n \)
- \( \tan x = 1 \) at \( x = n\pi + \frac{\pi}{4} \) for integers \( n \)

Thus, the intervals where \( \tan x \) lies between -1 and 1 are:

\[ x \in \left( n\pi - \frac{\pi}{4}, n\pi + \frac{\pi}{4} \right) \quad \text{for all integers } n \]

**Hint:** Look for the intervals where the tangent function is bounded between -1 and 1.

### Step 3: Determine the domain of \( f_3(x) = \sin(\sin^{-1} x) \)

The function \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, the domain of \( f_3(x) \) is:

\[ -1 \leq x \leq 1 \]

**Hint:** The inverse sine function has a limited range; remember its domain constraints.

### Step 4: Combine the domains

To find the overall domain of \( f(x) \), we need to take the intersection of the domains of \( f_1(x) \), \( f_2(x) \), and \( f_3(x) \):

1. Domain of \( f_1(x) \): \( \mathbb{R} \)
2. Domain of \( f_2(x) \): \( \bigcup_{n \in \mathbb{Z}} \left( n\pi - \frac{\pi}{4}, n\pi + \frac{\pi}{4} \right) \)
3. Domain of \( f_3(x) \): \( [-1, 1] \)

The intersection of these domains will be the values of \( x \) that satisfy all three conditions.

### Step 5: Identify the valid intervals

The only intervals from \( f_2(x) \) that fall within \( [-1, 1] \) are:

- For \( n = 0 \): \( \left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \)
- For \( n = 1 \): \( \left( \pi - \frac{\pi}{4}, \pi + \frac{\pi}{4} \right) \) which is outside of \([-1, 1]\)
- For \( n = -1 \): \( \left( -\pi - \frac{\pi}{4}, -\pi + \frac{\pi}{4} \right) \) which is also outside of \([-1, 1]\)

Thus, the only valid interval is:

\[ \text{Domain of } f(x) = \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]

**Final Answer:** The domain of the function \( f(x) \) is \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

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