Simple graphing problem giving me trouble

**kep84** · July 15th 2007, 11:39 AM

Hi all. Simple problem here, but I'm having a hard time for some reason. It's a graphing problem. The equation is y = sqrt(3) - 2 sin(x). The graph is drawn but different points along the graph need to be named. There are 6 different points needed. They are (0,A) (B,0) (Pi/2,C) (D,0) (4Pi/3,E) and (3Pi/2,F)

Let me know if I need to explain better. A is obviously sqrt(3) right? I'm having trouble with the rest of the answers. I would like to know why the answers are what they are, and not just the answers. Thanks for your time.

**Soroban** · July 15th 2007, 12:09 PM

Hello, kep84!

The equation is; . $y \:= \:\sqrt{3} - 2\sin(x)$

The graph is drawn but different points along the graph need to be named.

There are 6 different points needed.
They are: . $(0,\,A),\;(B,\,0),\;\left(\frac{\pi}{2},\,C\right) ,\;(D,\,0),\;\left(\frac{4\pi}{3},\,E\right),\;\le ft(\frac{3\pi}{2},\,F\right)$

You are right: it is a simple problem.

We are given: . $y \:=\:\sqrt{3} - 2\sin(x)$

Given the point: . $(0,\,A)$

What exactly is the problem?
. . It says: $x = 0$ . .Find $y$ .
Simple, right? . . . $y \:=\:\sqrt{3} - 2\sin(0) \:=\:\sqrt{3} - 0 \:=\:\sqrt{3}$
. . Therefore: . $A \,=\,\sqrt{3}$

Given the point: . $(B,\,0)$
. . It says: . $y \,=\,0$ . .Find $x$ .
We have: . $0 \:=\:\sqrt{3}-2\sin(x)\quad\Rightarrow\quad \sin(x) \:=\:\frac{\sqrt{3}}{2}\quad\Rightarrow\quad x \,=\,\frac{\pi}{3}$

Generalized: . $B \:=\:\begin{Bmatrix}\frac{\pi}{3} + n\pi \\ \frac{2\pi}{3} + n\pi\end{Bmatrix}$

Get the idea?

**kep84** · July 15th 2007, 12:16 PM

Yes. Thanks a bunch. It just clicked a minute ago, but thanks for the quick reply anyways. Cool forum here! Later.

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