Question

How do you evaluate the inverse sin of the sin of pi over 9?

`sin^-1((sin)pi/9)`

it's not on the unit circle, so i'm confused on where to start. help me out pls :)

Answer 1

`color(blue)(pi/9)`

Explanation:

If:

`y=sinx`

`arcsin(y)=x`

Hence:

`arcsin(sinx)=x`

So:

`arcsin(sin(pi/9))=pi/9`

The above makes sense if you think about what the inverse of a function is.

If we have an angle `x` and we put this through the function:

`f(x)=sinx`

we get a ratio returned.

When we put the ratio in the inverse function

`f^-1(x)=arcsin(x)`

we get back an angle.

In the problem we are given the angle `pi/9` and we got the ratio `y`, when we put `y` through the inverse function we got back `pi/9`

Answer 2

`pi/9 or 20^@`

Explanation:

`sin^-1((sin)pi/9)`

`:.pi/9 radians xx180/pi=20^@`

`:.=sin^-1((sin)20^@)`

`:.=sin^-1(0.342020143)`

`=20^@` or`pi/9`