Domain and range

hint : replace x with x+pi/2 . just one step away now :)

my domain is coming -1≤x≤+1 and range as [-2π/3,π/3]....someone verify....

Alright, if we say the domain is [-1,1] and with the definition

f(sin x) = x - pi/6, tell me what is f(1)?

BTW, I want to revise what I said in #4 to that the function itself cannot exist.

@ Debosmit : I really dunno.
@ hsbhatt sir : Now I understand sir.So if we imply a condition like say x is between [0,Ï€2] , we get a definite funcion.If this is the condition what should be the domain of the function....should it be [0,1] or [0,Ï€2] ????? Since we are finding the domain of f the latter may be the domain but it should also hold good for any value (sinx,cosx included) so it should be [0,1] ???????? Please help sir. Extremely confused. :(

But I think:

For x\in \left[\frac{\pi}{2}-1, 1 \right], the function f(x)= \sin^{-1} x-\frac{\pi}{6} works

@ Devil : Did u mean sin^-1 [x-π6] or sin^-1 x - π6....
If it is sin^-1 x - π6 even x=0 works I think.

I meant sin^-1x -Ï€6 and 0 does not work

given,

2f(sinx)+f(cosx)=x

putting ∩/2-x,in place of x,

2f(cosx)+f(sinx)=∩/2-x

On solving above two equations we get f(sinx)=x-∩/6

now putting sin^-1x in place of x we get

f(x)=sin^-1(x) - ∩/6

thus domain is [-1,1] (evident from above)

putting x=-1 we get f(x)=-2∩/3

and putting x=1 we get f(x)=∩/3

therefore, range is [-2∩/3.∩/3]

https://www.youtube.com/watch?v=CSjzlrL5hKA

https://www.youtube.com/watch?v=CSjzlrL5hKA