Thread: Graphing Trig Functions (can someone check these)

1. Graphing Trig Functions (can someone check these)

Okay so I have two problems i would like someone to check for me, i have to determine the amplitude, period, and phase shift of each function. Then i have to graph one period of the function. I don't know how to post my version of a graph here, but these are my calculations for the two equation i want checked so far.

Equation : y = sin (x - PI / 2)

Amplitude: 1

Period: 2 PI

Phase Shift : PI/2

Step: PI/2

X1: PI

X2: 3PI/2

X3: 5PI/2

X4: 7PI / 2

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Equation: y = sin (2x - PI / 2)

Amplitude: 1

Period: PI

Phase Shift: PI / 4

Step: PI / 4

x-values:

X1: PI / 4

X2: 3PI/4

X3: PI

X4: 5PI / 4

2. Originally Posted by MajorJohnson
Okay so I have two problems i would like someone to check for me, i have to determine the amplitude, period, and phase shift of each function. Then i have to graph one period of the function. I don't know how to post my version of a graph here, but these are my calculations for the two equation i want checked so far.

Equation : y = sin (x - PI / 2)

Amplitude: 1

Period: 2 PI

Phase Shift : PI/2

Step: PI/2

X1: PI

X2: 3PI/2

X3: 5PI/2 (This should be 2pi. Your increment is pi/2, which when added to your previous point 3pi/2, gives 4pi/2 = 2pi)

X4: 7PI / 2 (Likewise, this is incorrect)

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Notice the above edits.
Actually, I prefer to use FIVE points (where we include the endpoints of the period). In this case, we would "start" at
X0 = pi/2. Then
X1 = pi
X2 = 3pi/2
X3 = 2pi
X4 = 5pi/2

Then we look at sine and (after adjusting/scaling by... 1 in this case...) our points are
(X0, 0) = (pi/2, 0)
(X1, 1) = ...
(X2, 0)
(X3, -1)
(X4, 0)

3. free graphing program ... Graph

check your work yourself as you solve.

4. ^ Thanks for the software. This is what i needed.

Originally Posted by TheChaz
Notice the above edits.
Actually, I prefer to use FIVE points (where we include the endpoints of the period). In this case, we would "start" at
X0 = pi/2. Then
X1 = pi
X2 = 3pi/2
X3 = 2pi
X4 = 5pi/2

When i had PI / 2 under Step, i meant that as X0.

Then we look at sine and (after adjusting/scaling by... 1 in this case...) our points are
(X0, 0) = (pi/2, 0)
(X1, 1) = ...
(X2, 0)(X3, -1)
(X4, 0)
I just now noticed theres a pattern with the y-values for the sine curv ( This probably holds true in all cases for sine graphs) e. There are probably similar ones with the Cos curve. This should help me on my exam tomorrow. Thanks.