Question

sin inverse (sin 3π/5)

Answer 1

The value of sin^(-1) ( sin 3π/5 ) is supposed to be equal to 3π/5 
But actually it's not! 

For sin^(-1) ( sin x ) , x must be within [-π/2, π/2] 
Clearly 3π/5 > π/2 
So, 
sin^(-1) ( sin 3π/5 ) = sin^(-1) ( sin (π - 3π/5 )) = sin^(-1) ( sin 2π/5 ) = 2π/5 ∈ [-π/2, π/2] 

The answer is 2π/5

Answer 2

The value of sin^(-1) ( sin 3π/5 ) is supposed to be equal to 3π/5  

But actually it's not!  

For sin^(-1) ( sin x ) , x must be within [-π/2, π/2]  

Clearly 3π/5 > π/2  

So,  

sin^(-1) ( sin 3π/5 ) = sin^(-1) ( sin (π - 3π/5 )) = sin^(-1) ( sin 2π/5 ) = 2π/5 ∈ [-π/2, π/2]  

The answer is 2π/5