Pre-Calculus: Word Problems and Trig Equations

Great lesson, but could use more real application

07/25/2009

~ sharon5

When would you have a real problem where you would be looking for a position of a spring? I know that is what my students are going to be asking.

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Trigonometric Identities
Solving Trigonometric Equations
Solving Word Problems Involving Trigonometric Equations Page [1 of 2]
Now let’s take a look at an application where you actually use this trig business actually in practice. Now one thing
that a lot of physicists look at is what’s known as “spring motion.” This is where you have a spring on a surface, and
see if you were to pull it, it would start to go into a motion. Now it’s not going into a motion here because—well, it
stops because of the friction due to the table here and so forth. But you can imagine a frictionless surface, for
example if I had a whole bunch of holes drilled in here and I was blowing air up like one of those air hockey kinds of
things. It’s almost like no friction at all. If you were to start doing this kind of activity, it would start to wiggle back and
forth and the spring would keep going for a while and sort of oscillate. Well, that oscillatory movement actually can be
described using trig function. It turns out you can then model where the location of a particular mass is or a particular
weight is at any time just knowing how the motion goes, and that would be a trigonometric type of identity.
So let’s suppose for example that we have this spring motion in motion and it’s frictionless, on a frictionless surface,
so you’re playing this on sort of air hockey thing. And this thing is going back and forth. And suppose that we’re told
that the function that models the location of the mass, that blue blob, after T seconds is given by the following: l(T),
which is the location after T seconds, turns out—and there are trigonometric functions involved here:
sin 2T+ 3 cos 2T.
And now suppose I ask, okay, I want to find out all the times when this object is located where we started the
experiment. So I’m going to start the experiment right here. We start the experiment here, and so what does that
mean? It means that its position is, let’s say, at zero there. Let’s just call that zero. And now the thing starts to go,
and I want to know all the different times where that thing actually is in the zero position. So notice now, now, now,
now, now—do you see how it passed through that initial position a whole bunch of times? Let’s see if we can find all
the times where in fact that happens.
So what do I do? Well, I want to find out when this thing, at what times it produces the position of zero, zero position,
that’s where I start. So what I want o do is find all the T such that L equals zero. So I just set this equal to zero and
solve for T. Let’s do that. So that means that sin 2T= 3 cos 2T=0 . Well, how would you solve this? I guess one
thing you could do is bring this term to the other side. We’ll have two trig functions there and that’s a little bit
annoying. It becomes ? 3cos2T . And then what could I do? Well, then what I could do is divide both sides by
cos 2T . Now, why do I do that, by the way? What’s my thinking? What am I thinking here? Well, my thinking is that
I have two different trig functions that I have no idea how to grab them together, so that’s a little bit annoying. So how
do I get around that? Well, one way of getting around that is to say, hey, wait a minute, if I divide both sides by
cosine, then I see sine of something over cosine of something, and that’s tangent of something, and that actually
makes things a lot easier.
So I’m going to divide both sides by cosine. Of course, there is a worry. What if cosine were zero? Well, if cosine
were zero, then I know that in fact it’s not very likely that this term would be zero, and then that term can’t be zero
because sine and cosine can’t be zero at the same time. So I think we’re okay. I divide both sides by cosine, I see
sin 2 3
cos 2
T
T
= ? . Okay, great. Now, what is
sin
cos
? Well, that’s just tangent. So I see now that tan 2T = ? 3 . So
this whole question now boils down to finding the solutions to this slightly more simpler type of equation. So let’s see
if we can find those solutions.
So I want to find all the places where tan 2T = ? 3 . Well, remember what the tangent function looks like. I’ll sketch
it for you really fast. I’ll just sketch you part of it around the Y axis because we know that it repeats. So if I want to
find out what angle makes this ? 3 , that’s going to be down here somewhere. So it looks like it’s going to be sort of
a negative looking angle here. So let’s first of all figure out what the reference angle would be. So first of all, forget
about the negative sign. Suppose I just say to you, okay, I’m thinking of an angle and the tangent of it is 3 , what’s
the angle? Well, that’s one of those ones that hopefully is on that list of things that you remember. I happen to
Trigonometric Identities
Solving Trigonometric Equations
Solving Word Problems Involving Trigonometric Equations Page [2 of 2]
remember that that happens when the angle is 60° or
3
?
. And why? Because remember, tangent is
sin
cos
, and the
sine of 60° is—oh, wait a second. Wait a second, 60°? Maybe I don’t want 60°. No, I think I actually want—well, let’s
see. It’s either 60 or 30. Let’s work this out carefully.
So I have
sin
cos
?
?
and I want that to equal 3 . So now what do I want to go on here? What I want to have happen
is—let’s see. So if sin? were to be equal to
1
2
and cos? were to be equal to—okay, yes, I want
3
2
on top and
then
1
2
on the bottom. And you see how that would produce 3 ? Right. So in fact, my memory was okay. I
actually do want ? to be 60° because 60° gives me a sine of
3
2
and a cosine of
1
2
, and so the over-2’s cancel and
I’ll just get that. So in fact, it is in fact 60° or
3
?
.
Okay, so
3
?
. But that just gives me this solution here, but I want the negative of that. And so what I see is
2
3
T = ?? . So that’s that solution there. But of course there are infinitely many solutions, because I have a solution,
in fact, just one tangent sweep away, right? I’ve got a solution exactly here. How would I find that point? Well, I just
take this point and add ? to it. And I could find another one and another one and another one, so all I have to do is
actually add all the multiples of ? . So ? n where n=0,±1,±2... and so forth. So what does T equal? Well, I just
divide everything through by 2, so I see
6 2
?? +? n where n=0,±1,±2... and so on. So that’s an okay answer. But
of course, T is representing time in seconds, and so let’s just think about that. I can’t really have negative time, can I?
I wish I did. Then I could go backwards and get all those things done I’m supposed to. So if n were zero, that would
make negative time, so that’s no good. In fact, if n were any of these negative numbers, that would make negative
time. So in fact, what are the allowable n ’s? Well, if n were to equal 1, I don’t think that’s allowable because I think
that—well, let’s see.
2 6
? ?? is positive, so in fact, I’m allowed to equal 1, 2, 3, and anything bigger but I can’t be
allowed to equal any negative numbers because that would be negative time, and I’m not allowed to have zero
because that would be negative time.
So the answer would be at all these different times where this is in seconds. You can actually compute those times,
and those are all the times where in fact this mass is in that starting initial position, in the zero position. And again,
how did I find that? I just set up this equation, realized I could combine everything into one simpler type equation. I
solved that simple equation, found all the solutions this time, and then once I had all the solutions I had to make sure
they were all positive. Since this is actually an applied problem, I want to make sure that time is positive.
Anyway, you can see some applications to all this solving of these trig type equations. Messy but delicious. See you
soon.