Let `f: X rarr Y` be a function defined by f(x) = a sin ( x +`pi/4`) + c. If f is both one-one and onto, then find the set X and Y

Explanation

`f(x)=a "sin"(x+(pi)/(4)) + b cosx +c`
`=a{sin x "cos"(pi)/(4)+cosx "sin" (pi)/(4)}+b cosx +c`
`=(a)/(sqrt(2))sinx+((a)/(sqrt(2))+b)cos x+c " (1) " `
Let ` ((a)/(sqrt(2)))= r cos alpha, ((a)/(sqrt(2))+b)=r sin alpha`
` :. f(x)=r[cos alpha sinx+sin alpha cos x]+c`
`=r[sin(x+alpha)]+c`
where ` r=sqrt(a^(2)+sqrt(2)ab+b^(2))`
`and alpha= tan^(-1)((a+bsqrt(2))/(a)) " (2)" `
For `f` to be one-one, we must have `-pi//2 le x+ alpha le pi//2.` Thus,
domain is `[(-pi)/(2)-alpha, (pi)/(2)-alpha]` and range is `[c-r, c+r]`
`or X-=[(-pi)/(2)-alpha, (pi)/(2)-alpha] and Y-=[c-r,c+r]`