Question

Explain why inverse sine left parenthesis sine left parenthesis Start Fraction 3 pi over 4 End Fraction right parenthesis right parenthesis not equals Start Fraction 3 pi over 4 End Fraction when y = sin x and y equals inverse sine x are inverses.
(4 points)

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1 answer

To understand why \( \sin^{-1}(\sin(\frac{3\pi}{4})) \neq \frac{3\pi}{4} \), we need to consider the properties of the sine function and its inverse, the arcsine function (or inverse sine function).

  1. Sine Function Behavior: The sine function, \( y = \sin(x) \), is periodic and oscillates between -1 and 1. Its values repeat every \( 2\pi \). In the range of \( x \) from \( 0 \) to \( 2\pi \), the sine values rise from 0 to 1 in the first quadrant, then fall back to 0 in the second quadrant, go negative in the third quadrant, and return to 0 in the fourth quadrant.

  2. Range of Inverse Sine: The inverse sine function, \( y = \sin^{-1}(x) \), is defined to take values in the range of \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\). That means for any input \( x \) in the domain of \([-1, 1]\), the output of \( \sin^{-1}(x) \) must fall within those bounds.

  3. Calculating \( \sin(\frac{3\pi}{4}) \):

    • We find \( \sin(\frac{3\pi}{4}) \) which equals \( \sin(180^\circ - 45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \).
    • So, \( \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} \).
  4. Applying Inverse Sine: Next, we apply the inverse sine to this result: \[ \sin^{-1}(\sin(\frac{3\pi}{4})) = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \] The angle whose sine is \( \frac{\sqrt{2}}{2} \) that falls within the range of the inverse sine function is \( \frac{\pi}{4} \) (or \( 45^\circ \)).

  5. Conclusion: Therefore, \[ \sin^{-1}(\sin(\frac{3\pi}{4})) = \frac{\pi}{4} \neq \frac{3\pi}{4}. \] This is because while \( \frac{3\pi}{4} \) has the same sine value as \( \frac{\pi}{4} \), it does not fall within the restricted range of the arcsine function. The important point illustrates how the inverse sine function can only return angles within \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

In summary, the fact that \( \sin^{-1}(\sin(x)) \) does not return \( x \) for \( x \) outside the principal range of the inverse sine highlights the key relationship between a function and its inverse: they must be defined over compatible domains and ranges to satisfy the property of being true inverses.