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.

Answer 1

To understand why \(\sin^{-1}(\sin(\frac{3\pi}{4})) \neq \frac{3\pi}{4}\), we need to examine the definitions of the sine function and its inverse, the arcsin (or inverse sine) function, as well as the restrictions placed on the arcsin function.

1. **Understanding the Sine Function:**
The sine function, \(y = \sin(x)\), is periodic with a range from -1 to 1. It is defined for all real numbers \(x\), but it is not one-to-one over its full domain. One complete cycle of \( \sin(x) \) occurs between \(0\) and \(2\pi\), where each value in the range from \(-1\) to \(1\) is achieved multiple times.

2. **Understanding the Inverse Sine Function:**
The arcsine function, \(y = \sin^{-1}(x)\), is defined to be the inverse of the sine function but is restricted to a specific domain in order to make it one-to-one. The typical range for \(y = \sin^{-1}(x)\) is \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\). This means that the arcsine function will only return values within this interval.

3. **Evaluation of \(\sin(\frac{3\pi}{4})\):**
We compute:
\[
\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.
\]

4. **Applying the Inverse Sine Function:**
Now, we take the sine of the angle:
\[
\sin^{-1}\left(\sin\left(\frac{3\pi}{4}\right)\right) = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right).
\]
Since \(\frac{\sqrt{2}}{2}\) corresponds to the sine of both \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\), we must consider the output of the arcsin function. Because the output of \(y = \sin^{-1}(x)\) is restricted to \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\), the value returned will be:
\[
\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4},
\]
not \(\frac{3\pi}{4}\).

5. **Conclusion:**
Therefore, we conclude that:
\[
\sin^{-1}(\sin(\frac{3\pi}{4})) \neq \frac{3\pi}{4},
\]
because \(\sin^{-1}(x)\) gives the principal value of the angle whose sine is \(x\), and this output must fall within the defined range of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

In summary, the reason \(\sin^{-1}(\sin(\frac{3\pi}{4}))\) does not equal \(\frac{3\pi}{4}\) is due to the range restrictions of the inverse sine function, which only output values from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).