Draw the graph of `y=sin^(-1)|sin x|and y=(sin^(-1)|sinx|)^(2),0lexle2pi`

Explanation

To solve the problem of drawing the graphs of \( y = \sin^{-1} |\sin x| \) and \( y = (\sin^{-1} |\sin x|)^2 \) for \( 0 \leq x \leq 2\pi \), we will break it down into steps.

### Step 1: Understanding the function \( y = \sin^{-1} |\sin x| \)

1. **Domain and Range**:
- The function \( \sin x \) oscillates between -1 and 1. Therefore, \( |\sin x| \) will oscillate between 0 and 1.
- The range of \( \sin^{-1} \) is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). However, since we are taking the absolute value, the range of \( \sin^{-1} |\sin x| \) will be from 0 to \( \frac{\pi}{2} \).

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