Why doesn’t sin^-1(sin(3pi/4)) not equal 3pi/4 when y=sin x and y+sin^-1x are inverses?
Is it because its an odd function?
The Correct Answer and Explanation is :
The reason why (\sin^{-1}(\sin(\frac{3\pi}{4}))) does not equal (\frac{3\pi}{4}) is not because the sine function is odd, but because the range of the inverse sine function, (\sin^{-1}), is limited to ([- \frac{\pi}{2}, \frac{\pi}{2}]). This means that (\sin^{-1}) only outputs values within this interval.
The sine function, (\sin(x)), is indeed an odd function, meaning that (\sin(-x) = -\sin(x)). However, the key aspect affecting the inverse sine function is its principal value range, which is chosen to ensure that the function is one-to-one and thereby invertible.
When you calculate (\sin(\frac{3\pi}{4})), you get (\frac{\sqrt{2}}{2}). The angle (\frac{3\pi}{4}) (or 135 degrees) lies in the second quadrant of the unit circle, where sine values are positive. However, because the output of (\sin^{-1}) must fall within ([- \frac{\pi}{2}, \frac{\pi}{2}]), and (\frac{3\pi}{4}) does not fall within this interval, (\sin^{-1}(\sin(\frac{3\pi}{4}))) does not return (\frac{3\pi}{4}). Instead, it returns (\frac{\pi}{4}), which is the angle in the principal value range whose sine is also (\frac{\sqrt{2}}{2}).
This demonstrates an important property of inverse trigonometric functions: they return the angle from the principal value range that corresponds to the given sine value, not necessarily the original angle used in the sine function. This distinction is crucial for understanding how inverse functions are defined and operate within their restricted domains and ranges.
I’ll generate the image to illustrate this concept visually now.
Here’s the illustration showing the unit circle with angles (\frac{3\pi}{4}) and (\frac{\pi}{4}). It highlights their sine values and the principal value range of the inverse sine function. This visual should help clarify why (\sin^{-1}(\sin(\frac{3\pi}{4}))) returns (\frac{\pi}{4}) instead of (\frac{3\pi}{4}). You can see how the range of the inverse sine function restricts its output to the first and fourth quadrants.
