Question #7a7a9

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Question #7a7a9

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Answer 1 · 2017-12-05T05:29:04

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

Explanation:

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

$sin (0 - k) + c = 0$ $sin(0-k)+c = 0$
$sin (\frac{4 π}{3} - k) + c = 0$ $sin((4pi)/3-k)+c = 0$

Subtracting one from another, we get a single equation with a single unknown $k$ $k$ :
$sin (- k) - sin (\frac{4 π}{3} - k) = 0$ $sin(-k) - sin((4pi)/3-k) = 0$

This canbe simplified as
$- sin (k) - sin (\frac{4 π}{3}) \cdot cos (k) + cos (\frac{4 π}{3}) \cdot sin (k) = 0$ $-sin(k) - sin((4pi)/3)*cos(k)+cos((4pi)/3)*sin(k) = 0$

Since $sin (\frac{4 π}{3}) = sin (\frac{π}{3} + π) = - sin (\frac{π}{3}) = - \frac{\sqrt{3}}{2}$ $sin((4pi)/3) = sin(pi/3+pi) = -sin(pi/3) = -sqrt(3)/2$
and $cos (\frac{4 π}{3}) = cos (\frac{π}{3} + π) = - cos (\frac{π}{3}) = - \frac{1}{2}$ $cos((4pi)/3) = cos(pi/3+pi) = -cos(pi/3) = -1/2$ ,
our equation looks like

$- sin (k) + \frac{\sqrt{3}}{2} \cdot cos (k) - \frac{1}{2} \cdot sin (k) = 0$ $-sin(k)+sqrt(3)/2*cos(k)-1/2*sin(k) = 0$
Simplifying it to
$\frac{\sqrt{3}}{2} \cdot cos (k) = \frac{3}{2} \cdot sin (k)$ $sqrt(3)/2*cos(k) = 3/2*sin(k)$
or
$\frac{\sqrt{3}}{3} \cdot cos (k) = sin (k)$ $sqrt(3)/3*cos(k) = sin(k)$

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

$f (x) = sin (x - \frac{π}{6}) + \frac{1}{2}$ $f(x) = sin(x-pi/6)+1/2$

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Question #7a7a9

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