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

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?

• How, according to you, is the inverse sine function defined? – imranfat Nov 30 '16 at 21:54
• @imranfat As the inverse to the sine function. – Al Jebr Nov 30 '16 at 21:55
• $\sin$ isn't injective. Hence it has no global inverse. The principal branch has values in $[-\pi/2,\pi/2]$. If you take the branch of $\arcsin$ with values in $[\pi/2, 3\pi/2]$, you will get $\arcsin (\sin 5\pi/8)) = 5\pi/8$. But then $\arcsin (\sin (3\pi/8)) \neq 3\pi/8$. – Daniel Fischer Nov 30 '16 at 21:57
• There is no such thing as "the" inverse. Since $\sin$ is not 1-1 (and in fact is countably-infinite-to-1), many inverses are possible. The inverse you have in mind does not match the inverse the calculator has in mind. – MPW Nov 30 '16 at 21:57
• @AlJebr The arcsine function is the inverse of the sine function only under certain "restrictions". I think Ian's answer is a good one. In short, the inverse of a sine is truly only defined for the sine function on interval $[-90,90]$ (or in radian equivalence) – imranfat Nov 30 '16 at 22:06

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 $\frac{3\pi}{8}$.

• Is there a value I can subtract from the angle in the second quadrant to get its respective angle in the first quadrant, in general? For instance, in this case it would be $2\pi/8$. But what about some other angle in the second quadrant? – Al Jebr Dec 2 '16 at 1:36

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.

• @KitterCatter Done. – Ian Nov 30 '16 at 22:09
• Wonderful!!!! Thanks – Kitter Catter Nov 30 '16 at 22:13

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 $\theta = \frac{3\pi}{8}$.