# Could really use some help

• Apr 11th 2006, 11:05 AM
mazdatuner
Could really use some help
Im at a point in my book where I have to graph functions of sine and cosine
one such example of the problems is

Graph 1 cycle and lable 5 points on each.
F(x)=sin pie x <---- that is "F of X with sine...then the symbol for pie and x next to it

I know the points and can lable the amplitude and period and I can even graph it on my TI-83 but I am lost when It comes to graphing it by hand ....any help would be great :)

another question is
Solve for all angles over the interval [0,2pie] such that Sin of theta = 1/2

I have some Ideas and have read the book but I still dont know where to even start with this one either :(
• Apr 11th 2006, 03:04 PM
ThePerfectHacker
Quote:

Originally Posted by mazdatuner
Im at a point in my book where I have to graph functions of sine and cosine
one such example of the problems is

Graph 1 cycle and lable 5 points on each.
F(x)=sin pie x <---- that is "F of X with sine...then the symbol for pie and x next to it

I know the points and can lable the amplitude and period and I can even graph it on my TI-83 but I am lost when It comes to graphing it by hand ....any help would be great :)

Try doing this, you need to graph $f(x)=\sin (\pi x)$.
You know you amplitude it 1 and your period is $\frac{2\pi}{|\pi|}=2$. On this interval $[0,2]$ you midpoint is a zero, meaning $x=1$ then, $f(x)=0$. And the mipoints of those, (the quarters of $[0,2]$) are the max and min (with amplitude 1), thus at $x=1/2$ and $x=3/2$ you have a max and min resprectively. Then, you can draw a responable graph knowing this.
• Apr 11th 2006, 03:07 PM
ThePerfectHacker
Quote:

Originally Posted by mazdatuner
Solve for all angles over the interval [0,2pie] such that Sin of theta = 1/2

First you know that $\sin (\pi/6)=1/2$.
But, you still need other solutions on interval $[0,2\pi]$.
Remember that in the 2nd quadrant the sine is positive. Thus, the other solution appears in the second quadrant. Which is $\pi-\pi/6=\frac{5\pi}{6}$.
Hence, the two solutions are,
$\theta=\left\{\begin{array}{c}\pi/6\\ 5\pi/6$