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Graphing Trigonometric Functions (page 1 of 3)

Sections: Introduction, Examples with amplitude and vertical shift, Example with phase shift

You've already learned the basic trig graphs. But just as you could make the basic quadratic x2, more complicated, such as –(x + 5)2 – 3, so also trig graphs can be made more complicated. We can transform and translate trig functions, just like you transformed and translated other functions in algebra.

Let's start with the basic sine function, f(t) = sin(t). This function has an amplitude of 1 because the graph goes one unit up and one unit down from the midline of the graph. This function has a period of because the sine wave repeats every units. The graph looks like this:

graph of sine wave, from -2pi to +4pi

Now let's look at g(t) = 3sin(t):

graph of 3*sin(t), showing same x-intercepts, but three times the height

Do you see that the graph is three times as tall? The amplitude has changed from 1 to 3. This is always true: Whatever number A is multiplied on the trig function gives you the amplitude; in this case, that number was 3. So 0.5cos(t) would have an amplitude of 1/2, and –2cos(t) would have an amplitude of 2 and would also be flipped upside down.

Now let's look at h(t) = sin(2t): Copyright © Elizabeth Stapel 2010 All Rights Reserved

graph of sin(2t), showing period is half as long because sine changes twice as fast

Do you see that the graph is squished in from the sides? Do you see that the sine wave is cycling twice as fast, so its period is only half as long? This is always true: Whatever value B is multiplied on the variable, you use this value to find the period ω (omega) of the trig function, according to this formula:

    general period formula:

    omega = <regular period> / |B|

For sines and cosines (and their reciprocals), the "regular" period is , so the formula is:

    period formula for sines & cosines:

    (omega) = 2(pi)/abs(B)

For tangents and cotangents, the "regular" period is π, so the formula is:

    period formula for tangents & cotangents:

    omega = pi / |B|

In the sine wave graphed above, the value of B was 2. (Sometimes the value of B inside the function will be negative, which is why there are absolute-value bars on the denominator.) The formula for sines and cosines says that cos(3t) will have a period of (2π)/3 = (2/3)π; on the other hand, the formula for tangents and cotangents says that tan(t/2) will have a period of (2π)/(1/2) = 4π.

(Note: Different books use different letters to stand for the period formula. In your class, use whatever your book or instructor uses.)

Now let's looks at j(t) = sin(t – π/3):

graph of sin(t - pi/3) in blue, with graph of sin(t) in gray, showing shift to the right by pi/3 units

Do you see that the graph (in blue) is shifted over to the right by π/3 units from the regular graph (in gray)? This is always true: If the argument of the function (the thing you're plugging in to the function) is of the form (variable) – (number), then the graph is shifted to the right by that (number) of units; if the argument is of the form (variable) + (number), then the graph is shifted to the left by that (number) of units. So cos(t + π/4) would be shifted to the left by π/4 units, and tan(t – 2π/3) would be shifted to the right by (2/3)π units. This right-or-left shifting is called "phase shift".

Now let's look at k(t) = sin(t) + 3:

graph of sin(t) + 3, showing sine wave moved three units upward




Do you see how the graph was shifted up by threeunits? This is always true: If a number D is added outside the function, then the graph is shifted up by that number of units; if D is subtracted, then the graph is shifted down by that number of units. So cos(t) – 2 is the regular cosine wave, but shifted downward two units; and tan(t) + 0.6 is the regular tangent curve, but shifted upward by 6/10 of a unit.

Putting it all together, we have the general sine function, F(t) = Asin(Bt – C) + D, where A is the amplitude, B gives you the period, D gives you the vertical shift (up or down), and C is used to find the phase shift. Why don't you always just use C? Because sometimes more is going on inside the function. Remember that the phase shift comes from what is added or subtracted directly to the variable. So if you have something like sin(2t – π), the phase shift is not π units! Instead, you first have to isolate what's happening to the variable by factoring: sin(2(t – π/2)). Now you can see that the phase shift will be π/2 units, not π units. So the phase shift, as a formula, is found by dividing C by B.

    For F(t) = Af(Bt – C) + D, where f(t) is one of the basic trig functions, we have:

    • A: amplitude is A
    • B: period is (2π)/|B|
    • C: phase shift is C/B
    • D: vertical shift is D

Let's see what this looks like, in practice, because there's a way to make these graphs a whole lot easier than what they show in the book....

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Cite this article as:

Stapel, Elizabeth. "Graphing Triogonometric Functions." Purplemath. Available from Accessed


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