Pretty simple Q tbh (1 Viewer)

[Ikki]

By definition, isn't an inverse function:


So sin inverse of sin 1=1
and sin inverse of sin 1.5=1.5

BUT, sin inverse of sin 2=pi-2
and sin inverse of sin3=pi-3

And even to the extent that: sin inverse of sin 100=100-32pi.

Noting that everything is in radians. Can someone explain this?

 

[seanieg89]

By definition, isn't an inverse function:


So sin inverse of sin 1=1
and sin inverse of sin 1.5=1.5

BUT, sin inverse of sin 2=pi-2
and sin inverse of sin3=pi-3

And even to the extent that: sin inverse of sin 100=100-32pi.

Noting that everything is in radians. Can someone explain this?

There is a little more to it.

1. When you talk about a function, you must also specify a domain X and a co-domain Y (we write f:X->Y. this means that we can apply f to all elements x of the set X and each one gives us an element f(x) of Y). So the function f(x)=x^2 defined on the positive reals is different to the function g(x)=x^2 defined on the whole real line, despite them agreeing on the positive reals.

2. An inverse of a function f:X->Y is a function g:Y->X such that g(f(x))=x for all x in X and f(g(x))=x for all x in Y.

By this definition of invertibility (which is the standard definition), the function sin(x) is invertible only if we choose X to be a set like [-pi/2,pi/2] (which would give Y=[-1,1] and g as the inverse sine function).

The inverse sine function is definitely NOT an inverse for sin(x) with domain R, as your work demonstrates.

 

[Ikki]

Alright , thanks alot