Question

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

Answer 1

Inverses cancel each other out. `sin^(-1)(x)` is just another way of writing an inverse, or `arcsin(x)`.

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

`color(blue)(arcsin(sin(2^@)) = 2^@)`

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(sin(114.59^@))`

The `sin(114.59^@)` evaluates to about `0.9093`, and the `arcsin` of that would then be `1.14159cdots`, i.e.

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

---

Note that this is NOT:

`1/(sin(sin2))`

which is not the same thing. If you did have `1/(sin(sin(2))`, it would be equal to `(sin(sin2))^(-1)`.

However, even though `sin^2(x) = (sinx)^2`, it does not mean that `sin^(-1)(x) = (sinx)^(-1)`.

Answer 2

Refer to the Explanation Section.

Explanation:

Recall the following Defn. of `sin^-1` fun.,

`sin^-1x=theta, |x| <=1 iff sintheta=x, theta in [-pi/2,pi/2].`

Substituting the value `x=sintheta,` recd. from the R.H.S., into

the L.H.S., we get,

`sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]..........(star)`

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^@,` or `2" radian."`

If it is `2^@,`then, it follows from `(star)` that,

`sin^-1(sin2^@)=2^@.`

In case, it is `2" radian,"` we note that,

`sin2=sin(pi-(pi-2))=sin(pi-2),`

where, since `(pi-2) in [-pi/2,pi/2],` we have, by `(star),`

`sin^-1(sin2)=pi-2.`