Why is $\sin^{-1}(\sin(\frac{5\pi}{8}))\ne \frac{5\pi}{8}$?

Question

I am defining a function and making sure that it works. I thought $\sin^{-1}(\sin(x))=x$ but if I put it into a calculator I get $\sin^{-1}(\sin(\frac{5\pi}{8}))\approx1.178$; which is not $\frac{5\pi}{8}\approx1.96$.

What is the reason for this?

Answer 1

If you translate the statement what is $\sin^{-1}(\sin(\frac{5\pi}{8}))$ into a statement in English it would say the following:

What angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ has the same sine value as the angle $\frac{5\pi}{8}$?

If you draw the unit circle and the angle $\frac{5\pi}{8}$ and then draw a horizontal line through the point where the angle intersects the unit circle, I believe you will see that the horizontal line also intersects the unit circle in quadrant I at the point of intersection with the angle $\frac{3\pi}{8}$. So the answer is $\frac{3\pi}{8}$.

Answer 2

Sine is not truly an invertible function. Calculators, when asked for $\sin^{-1}(y)$ with $y \in [-1,1]$, are usually designed to give (an approximation to) the (unique) solution of the equation $\sin(x)=y$ within the interval $[-\pi/2,\pi/2]$. In mathematical parlance this $x$ is usually referred to as $\arcsin(y)$, rather than $\sin^{-1}(y)$. Anyway, $5\pi/8$ is not in this interval, so when you ask for $\sin^{-1}(\sin(5\pi/8))$, you get (an estimate of) $3\pi/8$ instead.

Here is a plot of one period of $\sin$ along with the line $y=\sin(5\pi/8)$ to demonstrate the point. You were expecting the calculator to give you the intersection on the right, but actually it gave you the one on the left.

Answer 3

The sine function is not one-to-one on $\mathbb{R}$, so it has no inverse on $\mathbb{R}$. You can only consider its inverse on a domain that is one-to-one. The typical convention is to choose $[-\pi/2,\pi/2]$. So when your calculator computes $\sin^{-1}(\sin(\frac{5\pi}{8}))$, it will give an angle $\theta \in [-\pi/2,\pi/2]$ such that $\sin\theta = \sin(\frac{5\pi}{8})$. The value is $\theta = \frac{3\pi}{8}$.

Answer 4

$T(T^{-1}(x))=x$ is always true, for all x belonging to domain of $T^{-1}(x)$. $T$ represnts any Trigonometric function.

But, $(T^{-1}T(x))=x$ is not always true. It is only true when x belongs to the range of $T^{-1}(x)$. Range of $\sin^{-1}(x)$ is $[\frac{-\pi}{2},\frac{\pi}{2}].$