1Well I replaced x by pi/2 - x to get f(sinx) + f(cosx)=pi/6. Solving I get f(sinx) = x - pi/6. So is the domain all real values or range of sinx (i.e. -1≤x≤1) ????? Please reply.
If 2f(sinx)+f(cosx) = x for all real x, find the domain and range.
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1my domain is coming -1≤x≤+1 and range as [-2π/3,π/3]....someone verify....
341Alright, 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.
1@ 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. :(
11But I think:
For x\in \left[\frac{\pi}{2}-1, 1 \right], the function f(x)= \sin^{-1} x-\frac{\pi}{6} works
1@ 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.
1given,
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]