for part 'b' i got $\displaystyle f(x) = A+4cos(x +45) $Given that $\displaystyle f(x) = A + 2\sqrt{2}cosx -2\sqrt{2}sinx $

$\displaystyle -180\leq x \leq 180 $

where A is a constant,

(b) show that f(x) can be expressed in the form

$\displaystyle f(x) = A + R cos (x + \alpha) $

where R > 0 and 0 < α < 90,

(c) state the value of A,

But I dont know how to work out the value of 'A'? Can someone please explain, how to?

Thanks