Welcome back! When last we met we had just graphed a sine function - today we'll look at the anatomy of the sine wave and how to express it in a simple equation. The example above is the sine wave I graphed yesterday, and its equation is below it:
f(x) = sin(x)
Straightforward enough, but let's take a look at why. The equation of any sine function is written like this:
f(x) = A sin (P x)
Where A is the wave's amplitude, and P is the period. The period of the function is how many revolutions it completes in 2π radians. This got a little confusing for me, because no two textbooks seem to be able to agree on variables for the period or how you find it, but you can think of it this way - look for the point where our wave starts to repeat, and call that value |B| (that's the absolute value of B - so treat it as a positive number even if it's negative). Then, simply solve for 2π/|B|. This wave, since it's a sine wave, starts at zero, rises to its maximum, goes back through zero, into the negative, and then arrives at zero again at exactly 2π. So, given our formula, 2π/2π = 1, so the period is 1.
In the above example, the amplitude is also one (since the wave's range is between one and negative one - the measure of the amplitude is always half of the wave's total height on the graph - in this case, only the distance between the horizontal axis and the maximum value.
So, with an amplitude of one and a period of one, f(x) = 1sin(1x) = sin(x). Easy? Okay, let's look at a cosine wave.
f(x) = sin(x)
Straightforward enough, but let's take a look at why. The equation of any sine function is written like this:
f(x) = A sin (P x)
Where A is the wave's amplitude, and P is the period. The period of the function is how many revolutions it completes in 2π radians. This got a little confusing for me, because no two textbooks seem to be able to agree on variables for the period or how you find it, but you can think of it this way - look for the point where our wave starts to repeat, and call that value |B| (that's the absolute value of B - so treat it as a positive number even if it's negative). Then, simply solve for 2π/|B|. This wave, since it's a sine wave, starts at zero, rises to its maximum, goes back through zero, into the negative, and then arrives at zero again at exactly 2π. So, given our formula, 2π/2π = 1, so the period is 1.
In the above example, the amplitude is also one (since the wave's range is between one and negative one - the measure of the amplitude is always half of the wave's total height on the graph - in this case, only the distance between the horizontal axis and the maximum value.
So, with an amplitude of one and a period of one, f(x) = 1sin(1x) = sin(x). Easy? Okay, let's look at a cosine wave.
I pulled this, and the rest of the graphs in this post from the Khan Academy's modules on graphing trig functions, so if you want to try your hand, head over there and get your learnin' on.
So, here's a cosine wave. It's a little different from the sine, in that whereas the sine wave starts at zero, the cosine wave begins at its maximum value, so when you're looking for its period, you'll look for the point where it reaches its maximum again. The wave above, for instance, still has a period of one, since it completes a revolution at 2π. The amplitude, however, is different. Its maximum is at 2.5, so that's the amplitude - now, we like to express things in our function equations as fractions rather than decimals, and 2.5 as a fraction is 5/2 - five halves. So this graph would be expressed thus:
f(x) = (5/2)cos (x)
Let's look at a couple more examples.
So, here's a cosine wave. It's a little different from the sine, in that whereas the sine wave starts at zero, the cosine wave begins at its maximum value, so when you're looking for its period, you'll look for the point where it reaches its maximum again. The wave above, for instance, still has a period of one, since it completes a revolution at 2π. The amplitude, however, is different. Its maximum is at 2.5, so that's the amplitude - now, we like to express things in our function equations as fractions rather than decimals, and 2.5 as a fraction is 5/2 - five halves. So this graph would be expressed thus:
f(x) = (5/2)cos (x)
Let's look at a couple more examples.
Okay, so when we look at this, we can immediately see that it's a sine wave, and a non-shifted one at that (sometimes they will be shifted to the left or right of zero - the distance from zero is called the phase of the wave function). We can also see that the amplitude is 1.5, or 3/2 as a fraction. The wave begins to repeat at 1π, so if we solve for 2π/1π, we get 2, so the period is 2.
f(x) = (3/2)sin (2x)
Make sense? Here's another.
f(x) = (3/2)sin (2x)
Make sense? Here's another.
Okay - this is a cosine wave, clearly, since it begins at its maximum. The amplitude is one, since the maximum is one, but let's find the period. This wave repeats at 6π, so 2π/6π = 1/3. Thus:
f(x) = cos (1/3)x
Not too difficult, eh? I'll leave you with something fascinating to ponder - one wave function we haven't looked at yet is the tangent function. It does something interesting. Take a look:
f(x) = cos (1/3)x
Not too difficult, eh? I'll leave you with something fascinating to ponder - one wave function we haven't looked at yet is the tangent function. It does something interesting. Take a look:
Weird, huh? I'll leave you to ponder exactly why the tangent wave does this - but here's a hint. At 90 degrees (π/2 radians) and 270 degrees (3π/2 radians), the value of tan ɵ actually approaches infinity - that's why it spikes off the grid like that. If you can noodle out why that happens, leave a comment. I'll give you the answer in the next post.
Have a great weekend everybody!
Have a great weekend everybody!
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