# How do you calculate sin^-1(sin2)?

Jul 25, 2016

Inverses cancel each other out. ${\sin}^{- 1} \left(x\right)$ is just another way of writing an inverse, or $\arcsin \left(x\right)$.

Note that $\arcsin$ returns an angle, and if the angle is in degrees, then

$\textcolor{b l u e}{\arcsin \left(\sin \left({2}^{\circ}\right)\right) = {2}^{\circ}}$

If the $2$ is in radians, then in terms of degrees:

arcsin(sin(2 cancel"rad" xx 180^@/(pi cancel"rad"))) = arcsin[sin((360/pi)^@)]

$= \arcsin \left(\sin \left({114.59}^{\circ}\right)\right)$

The $\sin \left({114.59}^{\circ}\right)$ evaluates to about $0.9093$, and the $\arcsin$ of that would then be $1.14159 \cdots$, i.e.

color(blue)(arcsin(sin("2 rad")) = pi - 2 " rad").

Note that this is NOT:

$\frac{1}{\sin \left(\sin 2\right)}$

which is not the same thing. If you did have 1/(sin(sin(2)), it would be equal to ${\left(\sin \left(\sin 2\right)\right)}^{- 1}$.

However, even though ${\sin}^{2} \left(x\right) = {\left(\sin x\right)}^{2}$, it does not mean that ${\sin}^{- 1} \left(x\right) = {\left(\sin x\right)}^{- 1}$.

Aug 3, 2017

Refer to the Explanation Section.

#### Explanation:

Recall the following Defn. of ${\sin}^{-} 1$ fun.,

${\sin}^{-} 1 x = \theta , | x | \le 1 \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] .$

Substituting the value $x = \sin \theta ,$ recd. from the R.H.S., into

the L.H.S., we get,

${\sin}^{-} 1 \left(\sin \theta\right) = \theta , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \ldots \ldots \ldots . \left(\star\right)$

Now, regarding the Soln. of the Problem, we note that, there is

no mention about the Measure of the Angle $2 ,$ i.e., it is

not clear, it is ${2}^{\circ} ,$ or $2 \text{ radian.}$

If it is ${2}^{\circ} ,$then, it follows from $\left(\star\right)$ that,

${\sin}^{-} 1 \left(\sin {2}^{\circ}\right) = {2}^{\circ} .$

In case, it is $2 \text{ radian,}$ we note that,

$\sin 2 = \sin \left(\pi - \left(\pi - 2\right)\right) = \sin \left(\pi - 2\right) ,$

where, since $\left(\pi - 2\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] ,$ we have, by $\left(\star\right) ,$

${\sin}^{-} 1 \left(\sin 2\right) = \pi - 2.$