Sine function - getting coordinates from the graph.

Question

I have trouble with getting coordinates from the sine graph:

The format of the function is: $f(x)=\sin(x-k)+c$

I need to find the coordinates of A. Could you give me just a little hint where to start from? I understand that the period of the function is $2\pi$, and that the horizontal coordinate is $\frac{2x}{3}$, but I have no idea how to get the vertical coordinate which, according to the exercise, is $\frac{3}{2}$, all I can say is that it should be equal to $c+1$ as the maximum point is one because there is no vertical stretch, and the graph is translated up by $c$.

I am expected to solve the problem in this order.

Answer 1

We have $0=f(2 \pi)$ and $0=f(\frac{4 \pi}{3})$. This gives two equations for $k$ and $c$.

Solve this system of equations. Then look for $x_0 \in (0, \frac{4 \pi}{3}) $ such that $f'(x_0)=0$.

Then we have $A=f(x_0)$.

Answer 2

You have $$f(x)=\sin(x-k)+c$$ Using the two points $$f\left(\frac{4 \pi }{3}\right)=\sin\left(\frac{4 \pi }{3}-k\right)+c= 0\tag 1$$ $$f(2\pi)=\sin(2\pi-k)+c=-\sin(k)+c=0\tag 2$$

Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely $$\sin\left(\frac{4 \pi }{3}-k\right)+\sin(k)=0 $$ Use $\sin(p)+\sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .

On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=\cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.

Answer 3

I would start by figuring out $k$ and $c$. From the graph, you can see that $f(0)= \sin{-k} +c = - \sin{k} + c = 0 \Rightarrow c = \sin{k}$

and also

$ f\left(\frac{4\pi}{3} \right) = \sin{\left(\frac{4\pi}{3} - k \right)} + c = 0 $

Now we can use $\sin{(x-y)} = \sin{x} \cos{y} - \cos{x}\sin{y} $ $$ \Rightarrow \sin{\left(\frac{4\pi}{3} - k \right)} = \sin{\frac{4\pi}{3}} \cos{k} - \cos{\frac{4\pi}{3}}\sin{k} = -\frac{\sqrt{3}}{2} \cos{k} + \frac{1}{2}\sin{k} $$ and therefore $$ f\left(\frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2} \cos{k} + \frac{1}{2}\sin{k} + c = 0 $$ We can also use $\cos{k} = \sqrt{1- \sin^2{k}} = \sqrt{1-c^2}$, giving an equation for $c$: $$ -\frac{\sqrt{3}}{2} \sqrt{1-c^2} + \frac{1}{2}c + c = 0 $$ Can you continue from here?

Answer 4

• Symmetry gives the first negative minimum of $f(x)$ in the middle between $\frac{4\pi}{3}$ and $2\pi$ at $x_m = \frac{2\pi + \frac{4\pi}{3}}{2}= \frac{5}{3}\pi$. Since $\sin(\frac{3}{2}\pi)= -1$ we get the shift $$\frac{5}{3}\pi- k = \frac{3}{2}\pi \Rightarrow \boxed{k = \frac{\pi}{6}}$$
• $f(2\pi) = \sin(2\pi - \frac{\pi}{6}) + c = 0$ $$\Rightarrow \boxed{c = \sin \frac{\pi}{6} = \frac{1}{2} }$$