Question

I, sir!!-2

Answer 1

The inverse of function f:A→B exists if f is one-one onto i.e.,

y=f( x )⇒ f −1 ( y )=x .

The given inverse trigonometry function is sin −1 ( − 1 2 ) .

Let

sin −1 ( − 1 2 )=y ,

siny=− 1 2 =sin( −π 6 )

Since, the range of the principle value of sin −1 ( x ) ( −π 2 , π 2 ) .

− π 6 ∈( − π 2 , π 2 )

Thus, the principle value of sin −1 ( − 1 2 ) is − π 6 .