A constant function? i.e. f(x) =c .Pls verify.

put x= -x,

f(sin(-x)) = f(cos(-x)).

so, f(-Sinx)) = f(cosx)

we get, f(-Sinx)) = f(Sinx)

It implies, it can be true only when f(x) is constant, beacuse it's giving f(-1/2) =f(1/2) and f(-1) =f(1)

and so on...

Possible for constant function only. => f(x) =c.

@ Himanshu Jain

u got f(-x)= f(x)

f(x) is an even fn, not always a constant fn

Here i think  f(x) is not a unique function

f(sin x)= f(sin (pi/2 - x))

implies f(x) is symmetric about sin pi/4 and also - sin pi/4 ( as f is even)

i.e. 1/root 2 and -1/root2

so f(x) is any even fn symm abt above pts. try finding examples

f(x) = a*|sin (pi x^2)| + b is one of possible ans. where a & b ar real constants.another possible answer is

f(x)= x^2 for |x|<= 1/root2

     = 1 - x^2 for 1/root2 <=|x|<=1

@Filokak: it's not f(-x) = f(x)

it's actually, f(g(-x)) = f(g(x)). So, you can't say, it's an even function.

Both are different things. Here, Sin and Cos can be treated as a different funtion.

Ok, try this way:

put x=x+pi

f(-Sinx) = f(-Cosx).

Also, we have, f(-Sinx) = f(Sinx) = f(Cosx)

=> f(Cosx) = f(-Cosx).

So, it can be said as constant function b'cos to satisfy the same values for Cos and Sin, there are some particular values are possible only.

@Himanshu Jain: Here both are satisfied i.e. f(g(-x))=f(g(x)) and f(-x)=f(x) .....

[u already have f(sinx)=f(-sinx)..............also note sin(-x)=sinx ...

Try to verify the fn below.....it clearly satisfies the given condition

f(x) = a*sin (pi x^2) + b

----  f(cosx) =a*[sin pi * (cosx)^2]+ b

                    =a*[sin pi( 1- (sinx)^2)] +b

                    =a*[sin{ pi - pi (sinx)^2}]+b

                     =a*[sin pi ( sin x)^2] +b

                      = f(sinx)