SOLUTION: Hi, I'm terribly stuck with this question: The function f is defined on the domain [0,pi] by f(x)=4cos(x)+3sin(x) a) express in the form Rcos(x-y), where 0 < y < (pi)/2 b) Hen

 
4cos(x)+3sin(x) = R()[4cos(x)+3sin(x)] = R[(cos(x)+sin(x)] =

 R·cos(x-y) = R·[cos(x)cos(y)+sin(x)sin(y)]

So

R[(cos(x)+sin(x)] = R·[cos(x)cos(y)+sin(x)sin(y)]

These will be equal if we choose y and R such that 

 = cos(y) and  = sin(y)

So we draw a right triangle with angle y, adjacent side 4, opposite side 3,
and hypotenuse R:



So we choose y = arctan() = 0.6435
and calculate R =  =  =  = 5

Therefore 

f(x) = 4·cos(x)+3·sin(x) = R·cos(x-y) = 5·cos(x-0.6435)

f(x) will have maximum value when cos(x-0.6435) has maximum value.
cos(q) has maximum value of 1 when q = 0, so we set

              x-0.6435 = 0
                     x = 0.6435
Edwin