Graphing Trig Functions

Factual Questions
1

I am in the middle of teaching a unit on graphing trigonometric functions and am getting hung up on an apparent anomaly.

We are graphing functions that have the following form:

y = C + AsinB(x - D) and y = C + AcosB(x - D)

where:[ul]
[li]A = Amplitude (Vertical Stretch)[/li][li]360/abs(B) = Period (Horizontal Stretch)[/li][li]C = Sinusoidal Axis (Vertical Shift)[/li][li]D = Phase Displacement (Horizontal Shift).[/li][/ul]
Since the period is defined using the absolute value of B, the graphs of these functions should be the same using +/- B so long as all other values remain the same.

This works for cosine graphs but not for sine graphs. Why?

An example is the following function group.

y1 = 3 + 4cos(8(x - 5))
y2 = 3 + 4cos(-8(x - 5))
y3 = 3 + 4sin(8(x - 5))
y4 = 3 + 4sin(-8(x - 5))

The graphs of y1 and y2 are identical but the graphs of y3 and y4 are mirror images of one another, reflected about the sinusoidal axis.

A great grapher can be found here but functions must be entered with * representing multiplication.

2

Cosine is an even function, meaning that cos(–x) = cos(x).
Sine is an odd function, meaning that sin(–x) = –sin(x).

Multiplying the argument of either sin or cos by –1 won’t affect the period (how long before the graph “repeats”). It will not affect the graph of cos at all, but it will affect the graph of sin.

3

I agree with Thudlow. Why do you think that “Since the period is defined using the absolute value of B, the graphs of these functions should be the same using +/- B…”? I think you could say the period should be the same in either case, but not the graph itself, which is not defined in terms of abs(B).

4

Thanks, guys. That makes perfect sense. I don’t know how I missed that.

:smack: :smack: