This is the question from my textbook:
The correct answer for the coordinates of $A$ is $\left(\frac{2\pi}{3},\frac{3}{2}\right)$. I've tried making three equations and solving by substitution for $k$ and $c$, but I do not get the correct values. Here's the equations I made:
$0=sin(\frac{4\pi}{3})+c$
$0=sin(\frac{2\pi}{3})+c$
$0=sin(2\pi)+c$
Correct value for $k$ is $\frac{\pi}{6}$, for $c$ it is $\frac{1}{2}$.
Help would be greatly appreciated. Thank you.
$\endgroup$ 23 Answers
$\begingroup$hint
from the graph, we derive that
$$f (0)=f (4\frac \pi 3)=0$$
thus
$$\sin (-k)+c=\sin (4\frac \pi 3-k)+c=0$$
from here, $$-k=\pi-(4\frac \pi 3-k) $$
You can finish.
$\endgroup$ $\begingroup$b) is easier than a) so I am going to start there.
the function is period with period $2\pi$
that is $f(x) = f(x+2n\pi)$
if $f(x) = 0$ then $f(2\pi) = 0,f(-2\pi) = 0$ etc.
In the interval $[-4\pi,0]$
$0-4\pi, \frac {4\pi}{3}-4\pi, -2\pi, \frac {4\pi}{3} - 2\pi, 0$
or, $-4\pi, -\frac {8\pi}{3}, -2\pi, -\frac {2\pi}{3}, 0$
c) Every period there are two solutions. There are two in the cycle $[0,2\pi)$ and two more in $[2\pi,4\pi)$, etc.
there are 8 solutions in $[0,8\pi)$
there is no solution in $[8\pi, 9\pi)$
a) Phase-shift:
The maximal and minimal values of $\sin(x-k)-c$ will be half-way between the zeros.
$x = \frac {2\pi}{3}, \frac {5\pi}{3}$
The maximal and minimal values of an un-shifted sin function are at $x= \frac {\pi}{2},x= \frac {3\pi}{2}$
the phase-shift is $\frac {\pi}{2} - \frac {2\pi}{3} = -\frac {\pi}{6}$
Vertical shift:
$\sin(0-\frac {\pi}{6}) + c = 0\\ -\frac 12 + c = 0$
$f(x) = \sin(x-\frac {\pi}{6}) + \frac 12$
and finally A
$f(\frac {2\pi}{3}) = \sin(\frac {\pi}{2}) + \frac 12 = \frac 32$
$\endgroup$ 1 $\begingroup$$\sin(2π-k)+c=0$
This is equivalent to $\sin(k)=c$
$\sin(\frac{4π}{3}-k)+c=0$
$-\frac{\sqrt{3}}{2}\cos(k)+\frac{1}{2}\sin(k)+\sin(k)=0$
$-\sqrt{3}\cos(k)=-3\sin(k)$
$\tan(k)=\frac{\sqrt{3}}{3}$
$k=\frac{π}{6}$
$c=\sin(\frac{π}{6})=\frac{1}{2}$
$\endgroup$ 0
