Question

Question #7a7a9

Answer 1

(a) (i) Point `A` (local maximum) is at coordinates `((2pi)/3, 3/2)`
......(ii) `f(x) = sin(x-pi/6)+1/2`
(b) Zero points are at `x=2piN`, where `N=-2, -1, 0` and `x=(4pi)/3+2piN`, where `N=-2, -1` (to be in the interval `[-4pi,0]`

Explanation:

Provided (from the picture of a graph) that `f(0)=0` and `f((4pi)/3)=0`, we can write the following system of two equations with two unknown `x` and `c`:

`sin(0-k)+c = 0`
`sin((4pi)/3-k)+c = 0`

Subtracting one from another, we get a single equation with a single unknown `k`:
`sin(-k) - sin((4pi)/3-k) = 0`

This canbe simplified as
`-sin(k) - sin((4pi)/3)*cos(k)+cos((4pi)/3)*sin(k) = 0`

Since `sin((4pi)/3) = sin(pi/3+pi) = -sin(pi/3) = -sqrt(3)/2`
and `cos((4pi)/3) = cos(pi/3+pi) = -cos(pi/3) = -1/2`,
our equation looks like

`-sin(k)+sqrt(3)/2*cos(k)-1/2*sin(k) = 0`
Simplifying it to
`sqrt(3)/2*cos(k) = 3/2*sin(k)`
or
`sqrt(3)/3*cos(k) = sin(k)`

It's easy to prove that `cos(k)!=0` (otherwise, the second equation in our system would not be satisfied.)
So, dividing by `cos(k)`, we get
`tan(k) = sqrt(3)/3`
One of the solution of this equation is `k=pi/6`, then from the first equation of our system `c=sin(pi/6)=1/2`.

`f(x) = sin(x-pi/6)+1/2`