## sin−1(sin(x))=__________for x in__________

### x; [−π/2,π/2]

The answer to this question depends on whether you are working in degrees or radians. I will provide answers for both cases:

If working in degrees:

sin−1(sin(x))=x for x in [-90°, 90°]

If working in radians:

sin−1(sin(x))=x for x in [−π/2, π/2]

Explanation:

The function sinx takes an angle as input and returns the sine of that angle. The inverse sine function, sin−1x (sometimes also written as arcsinx or asin), takes a value (between -1 and 1) as input and returns the angle whose sine is that value.

When you apply the inverse sine function to sinx, the result is simply the angle x itself, as long as x is within the range of possible angles whose sine is within the range of the inverse sine function.

In degrees, the range of the inverse sine function is from -90° to 90°, which corresponds to the range of possible values of sinx. Therefore, sin−1(sin(x))=x for x in [-90°, 90°].

In radians, the range of the inverse sine function is from -π/2 to π/2, which again corresponds to the range of possible values of sinx. Therefore, sin−1(sin(x))=x for x in [−π/2, π/2].

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