WEBVTT
00:00:07.103 --> 00:00:15.197
Today we're going to look at trigonometric equations. A trigonometric equation has a trig function as part
00:00:15.199 --> 00:00:24.039
of the equation. There are two types: an identity which we studied previously and conditional equations
00:00:24.041 --> 00:00:34.880
which we're going to study today. Let's review a minute. What is an identity? An identity is true for all
00:00:34.882 --> 00:00:41.630
solutions of the variable. Now I'm going to write two examples on the board, and I want you to be
00:00:41.632 --> 00:00:49.127
copying these because we're going to need them in some other examples. So the first one that you should
00:00:49.129 --> 00:01:00.131
recognize is the sine of theta squared plus the cosine of theta squared equals 1. This is our famous
00:01:00.133 --> 00:01:07.995
Pythagorean identity, and we will use it. The second one that I'm going to write here, be copying
00:01:07.997 --> 00:01:23.497
remember, is the cosine of 2 theta equals 2 cosine theta squared minus 1. So what is a conditional
00:01:23.499 --> 00:01:32.493
trigonometric equation? Where identities are true for all solutions of the variable, a conditional trigonometric
00:01:32.495 --> 00:01:41.997
equation is only true for specific values of the variable. The example that we're going to start with today is
00:01:41.999 --> 00:01:55.996
this one. 2 cosine theta plus 1 equals 0.
We will get started in just a minute.
00:01:58.290 --> 00:02:06.792
We're going to look at trigonometric equations that are linear in form. The example that we spoke about
00:02:06.794 --> 00:02:16.544
earlier is 2 cosine theta plus 1 equals 0, and we want to do two things from our directions. We want to
00:02:16.546 --> 00:02:25.546
determine a general formula for the solutions of the equation. We're going to let k represent any integer.
00:02:25.548 --> 00:02:35.754
Then as a b part, we're going to determine the specific solutions, if any, on the interval 0 to 2 pi.
00:02:35.756 --> 00:02:42.094
So in looking at this equation, the first thing you might ask yourself is, "What do I do with it? I don't know
00:02:42.096 --> 00:02:52.936
where to begin to solve this equation." So you might come here on the side and use substitution. You could
00:02:52.938 --> 00:03:04.439
use any letter, u or x or y or w, and let that be the trig function. Making the substitution, you should see
00:03:04.441 --> 00:03:13.188
an equation that you recognize from algebra. What do we do to solve this equation? Well, very simply,
00:03:13.190 --> 00:03:23.805
we subtract the 1, and then we divide by 2 on both sides of the equation. So in algebra, we would have this
00:03:23.807 --> 00:03:32.553
solution. So now what do we do with our trigonometric equation? The exact same thing. We're going to
00:03:32.555 --> 00:03:48.078
subtract 1 and then divide by 2. Ok, unlike the algebra equation, where we were finished, we now have
00:03:48.080 --> 00:04:01.076
to solve for the variable theta. We now want to find the general formula for all of the solutions. Well, in
00:04:01.078 --> 00:04:09.828
order to do that, we need to do a couple of things. The first thing we need to do is find the reference angle,
00:04:09.830 --> 00:04:22.330
theta r, so I'm going to locate. Let's remember our right triangles, and let's remember that the cosine function
00:04:22.332 --> 00:04:34.827
is the ratio of the adjacent side, so we're going to let 1 be the adjacent side, 2, ratio of the adjacent side
00:04:34.829 --> 00:04:46.573
over the hypotenuse. So we have 1 divided by 2. That means the opposite side is the square root of 3.
00:04:46.575 --> 00:05:02.076
Ok, what is that reference angle? Well, it is pi over 3. Now in this problem our ratio is negative, so pi
00:05:02.078 --> 00:05:12.745
over 3 is not one of our answers. [In] which two quadrants is the cosine function negative? Quadrant two,
00:05:12.747 --> 00:05:23.496
which will give us one solution, and quadrant three, which will give us a second solution. Ok, reference
00:05:23.498 --> 00:05:39.246
angle pi over 3. Now let's see what the general formula is. The angle one, quadrant two, we will find by
00:05:39.248 --> 00:05:58.454
subtracting the reference angle from pi. So we have pi minus pi over 3, which is 2 pi over 3, plus the
00:05:58.456 --> 00:06:11.206
period of our cosine function is 2 pi, and we will multiply that times k. This will give us a set of solutions.
00:06:11.208 --> 00:06:27.571
What about the second angle, theta two? Well, in quadrant three, theta two will be pi plus the reference angle,
00:06:27.573 --> 00:06:48.076
so that'll be pi plus pi over 3, which is 4 pi over 3, plus again 2 pi times k. That is the answer to the a
00:06:48.078 --> 00:06:58.329
part of our problem. What do we do for the b part? We want to find specific solutions on the interval
00:06:58.331 --> 00:07:11.101
from 0 to 2 pi. All we have to do, we're going to begin by letting k be 0. 0 times 2 pi of course is 0.
00:07:11.103 --> 00:07:26.102
That's going to give us a solution of 2 pi over 3. Same thing, let k be 0 here. That gives us 4 pi over 3.
00:07:26.104 --> 00:07:35.877
Ok, so do we let k be 1? Well, let's see what happens. If k is 1, then we would be adding 2 pi to each,
00:07:35.879 --> 00:07:49.124
2 pi over 3 and 4 pi over 3. Both of those answers are not in our interval, 0 to 2 pi. So we see that on that
00:07:49.126 --> 00:07:53.378
given interval we have two solutions.
00:07:55.560 --> 00:08:03.312
Let's look at this example. We want to determine a general formula for the solutions of the equation
00:08:03.314 --> 00:08:13.820
tangent theta plus 1 equals 0. In our formula we're going to let k represent any integer, and then in the
00:08:13.822 --> 00:08:22.247
b part, we're going to determine the specific solutions, if any, on the interval 0 to 2 pi. Now, this example
00:08:22.249 --> 00:08:32.017
is an example of a trigonometric equation that is linear in form. When we first look at it you may not
00:08:32.019 --> 00:08:44.862
be sure exactly what to do with it. So you can very easily come here to the side, let u be your trig function,
00:08:44.864 --> 00:08:57.954
tangent theta. Making the substitution, we have the equation u plus 1 equals 0 and solving we subtract 1 from
00:08:57.956 --> 00:09:07.384
both sides. We see that u is negative 1. We're going to do the exact same procedure for our trig equation.
00:09:07.386 --> 00:09:20.226
We're going to subtract 1, and then we want to find all the solutions of theta. The ratio is negative,
00:09:20.228 --> 00:09:27.729
but to work this problem, we're going to first find the reference angle. So let's come over here to the side
00:09:27.731 --> 00:09:46.252
and find theta r. In our right triangle, we're going to set up tangent is the ratio. Let's rewrite that here.
00:09:46.252 --> 00:09:56.435
Tangent of theta r equals positive 1. We want to know where the tangent is positive. So tangent is the
00:09:56.437 --> 00:10:08.530
ratio of the opposite side to the adjacent side. Those sides are each 1. By the Pythagorean theorem, the
00:10:08.532 --> 00:10:19.464
hypotenuse is now the square root of 2, and we recognize the reference angle to be pi over 4.
00:10:19.466 --> 00:10:31.798
Back to our problem. Tangent theta is negative 1. Ok, where is the tangent of theta negative? There are
00:10:31.800 --> 00:10:43.393
two quadrants in which the tangent function is negative. Quadrant two, we'll call it theta one. Quadrant
00:10:43.395 --> 00:10:57.572
four, we'll find that solution and call it theta two. To find the first angle in quadrant two, we need to
00:10:57.574 --> 00:11:16.328
subtract the reference angle from pi, so pi minus pi over 4 is 3 pi over 4, and then quadrant four, we want to
00:11:16.330 --> 00:11:36.786
subtract the reference angle from 2 pi. 2 pi minus pi over 4 is 7 pi over 4. So we come back here, and
00:11:36.788 --> 00:11:50.377
we want to write our general formula. We have 3 pi over 4 plus pi k because the period of the tangent
00:11:50.379 --> 00:12:05.877
function is pi, and 7 pi over 4 plus pi k. Ok, but looking at this, this doesn't seem very concise, and it's not
00:12:05.879 --> 00:12:20.245
because if we let k be 1 and add pi to 3 pi over 4 we end up with 7 pi over 4. So the better way to write the
00:12:20.247 --> 00:12:38.428
general formula is simply 3 pi over 4 plus pi k. That will be our answer for the a part, the general formula.
00:12:38.430 --> 00:12:56.698
Now in the b part, we will begin by letting k be 0. That of course gives us 3 pi over 4. Let k be 1. As we
00:12:56.700 --> 00:13:13.701
previously said, pi plus 3 pi over 4 is 7 pi over 4. But can we let k be 2? I don't think so. 2 pi plus 3 pi over 4
00:13:13.703 --> 00:13:28.697
is outside of our interval. We have 2 solutions, 3 pi over 4, 7 pi over 4. What you want to remember about
00:13:28.699 --> 00:13:36.851
this problem is that when you're solving with the tangent function you're going to work it very differently
00:13:36.853 --> 00:13:44.300
than if your trig equation has the sine or the cosine function.
00:13:46.900 --> 00:13:56.942
We're going to look at this example. The sine of 2 theta equals 1 over the square root of 2. Our directions
00:13:56.944 --> 00:14:07.083
tell us to find the specific solutions, if any, on the interval from 0 to 2 pi. Now, this trig equation is linear
00:14:07.085 --> 00:14:15.927
in form, but it's a little different than some of the others that you may encounter in your homework, and
00:14:15.929 --> 00:14:27.268
that's because there is a constant multiplied times our angle theta. So what we want to think about is what
00:14:27.270 --> 00:14:38.696
solutions of this angle, what solutions when we take the sine of that angle, will equal 1 over the square
00:14:38.698 --> 00:14:48.791
root of 2. Back here, at the side, we can find the reference angle, but you may know it immediately because
00:14:48.793 --> 00:14:57.881
1 over square root of 2 sine function is very ... It's in our family, and it's very obvious to us. Sine function,
00:14:57.883 --> 00:15:14.633
ratio of the opposite side to the hypotenuse, we see the reference angle is pi over 4. I am going to set this up
00:15:14.635 --> 00:15:24.132
using a general formula to begin with, even though the problem does not call for me to give that in my
00:15:24.134 --> 00:15:38.089
final answer. So we'll write 2 theta equals pi over 4 plus 2 pi k. Now that's only one quadrant in which the
00:15:38.091 --> 00:15:48.183
sine function is positive. Ok, let's think about this. [In] which other quadrant is the sine function positive?
00:15:48.185 --> 00:16:00.027
Quadrant two, correct. That will give us a second solution which we will find by subtracting the... oops
00:16:00.029 --> 00:16:24.276
watch the mistake here. Theta two is pi minus the reference angle, so pi minus pi over 4 which is 3 pi over 4.
00:16:24.278 --> 00:16:34.775
Now if we did not have this 2, we would be done as far as finding a general formula, but we have to deal
00:16:34.777 --> 00:16:45.324
with the 2. The 2 is multiplied times theta. We need to take it to the other side, so we're going to divide.
00:16:45.326 --> 00:16:58.914
It's very important that we divide all four terms by the 2. Pi over 4 divided by 2 is pi over 8. 2 pi k divided
00:16:58.916 --> 00:17:18.914
by 2 is pi k. 3 pi over 4 divided by 2 is 3 pi over 8, and again, 2 pi k divided by 2 is pi k. Now we can start
00:17:18.916 --> 00:17:35.667
finding our specific solutions. We let k be 0. That gives us pi over 8, then 3 pi over 8. Then we'll let k be 1,
00:17:35.669 --> 00:17:53.663
so we're adding pi over 8 plus pi which is 9 pi over 8. Let k again be 1. Add pi plus 3 pi over 8. That gives
00:17:53.665 --> 00:18:04.152
us 11 pi over 8, and of course we cannot let k be 2 because we'd be adding 2 pi to each of the angles
00:18:04.154 --> 00:18:15.150
which is outside of our interval. So we see in this example that we have 4 solutions.
00:18:16.583 --> 00:18:26.675
We're going to look at a trigonometric equation that is quadratic in form and attempt to solve it. Our
00:18:26.677 --> 00:18:36.767
directions tell us to find the specific solutions, if any, on the interval from 0 to 2 pi, and our example reads
00:18:36.769 --> 00:18:49.358
3 times the cosecant of theta squared minus 4 equals 0. I'm not sure how to begin the problem, so I'm
00:18:49.360 --> 00:19:01.859
going to come here to the side. Let u equal my trig function, cosecant theta. Making the substitution, I have
00:19:01.861 --> 00:19:19.474
3 u squared minus 4 equals 0. Then I recall from algebra, I would add the 4, divide by 3, and then one
00:19:19.476 --> 00:19:28.568
more step. To undo the square, I would take the square root on both sides, but I need positive and
00:19:28.570 --> 00:19:43.563
negative square root of 4/3, and of course this simplifies to plus or minus 2 divided by the square root of 3.
00:19:43.565 --> 00:20:06.027
I'm going to do the same thing over here. I'm going to add the 4, divide by 3, take the square root, and I
00:20:06.029 --> 00:20:24.460
end up with the cosecant of theta is plus or minus 2 divided by the square root of 3. I need to solve for theta.
00:20:24.462 --> 00:20:36.890
I'm going to find the reference angle. Here to the side, I'm going to find the reference angle such that the
00:20:36.892 --> 00:20:54.574
cosecant of the reference angle equals the positive ratio 2 divided by square root of 3. In quadrant one, my
00:20:54.576 --> 00:21:05.822
reference angle, I'm going to place the 2 and the square root of 3 at the appropriate sides. I know cosecant
00:21:05.824 --> 00:21:18.073
is the ratio, cosecant of theta is the ratio of the hypotenuse, so that's the 2, divided by the opposite side,
00:21:18.075 --> 00:21:32.194
which is square root of 3. That makes this side 1, and I know that my reference angle is pi over 3. Ok,
00:21:32.196 --> 00:21:41.536
that'll give me one solution. Let's write that solution down for positive 2 over square root of 3 in quadrant
00:21:41.538 --> 00:21:55.466
one, but consider this. Cosecant is positive in quadrant one. It's positive in quadrant two. It's negative in
00:21:55.468 --> 00:22:08.652
quadrant three and negative in quadrant four. So we have four solutions. Ok, pi over 3 is our solution
00:22:08.654 --> 00:22:20.900
in quadrant one. Let's call it theta one. Theta two, our second solution, we will find by subtracting the
00:22:20.902 --> 00:22:37.177
reference angle from pi which gives us 2 pi over 3, our second solution. Third solution, in quadrant three,
00:22:37.179 --> 00:22:57.951
we will find by adding pi to the reference angle. Pi plus pi over 3 is 4 pi over 3, third solution. Finally
00:22:57.953 --> 00:23:13.950
in quadrant four, we will get our solution by subtracting the reference angle from 2 pi,
00:23:13.952 --> 00:23:28.477
and that gives us 5 pi over 3. This trigonometric equation that is quadratic in form
00:23:28.479 --> 00:23:36.224
has four specific specific solutions on the interval from 0 to 2 pi.
00:23:38.069 --> 00:23:46.564
We'll take a look at this example. We want to determine the specific solutions, if any, on the interval from
00:23:46.566 --> 00:23:56.809
0 to 2 pi of the trigonometric equation that is quadratic in form, and our example reads 2 times the sine of
00:23:56.811 --> 00:24:10.311
theta squared plus the sine of theta equals 0. I'm going to make my substitution and let u be the sine of
00:24:10.313 --> 00:24:23.055
theta, so I can readily recognize how to solve this equation. In algebra, substituting u here, if u is the sine
00:24:23.057 --> 00:24:35.673
of theta, then sine of theta squared, will be u squared plus u equals 0. Then I recall from algebra when
00:24:35.675 --> 00:24:47.175
I have this on the left side of the equation I can factor the common factor, u, and after doing that, u times
00:24:47.177 --> 00:25:05.290
2u plus 1 equals 0. The two factors give me 0, so I can set each factor equal to 0 and solve independently.
00:25:05.292 --> 00:25:17.793
First solution is u equals 0. Second solution is u equals negative 1/2. I subtracted 1 then divided by 2.
00:25:17.793 --> 00:25:30.040
So that's what we want to do with this trig equation. The common factor is the sine of theta. Write that,
00:25:30.042 --> 00:25:45.004
then divide. Sine of theta into 2 sine squared is 2 sine theta plus sine of theta into sine theta is 1. Or you
00:25:45.006 --> 00:25:57.001
can simply look at what you have here and just substitute back. For u equals 0 we have sine theta equals 0,
00:25:57.003 --> 00:26:16.251
like this factor, and then 2 sine theta plus 1 equals 0. Solving this one, sine theta equals negative 1/2.
00:26:16.253 --> 00:26:29.957
We have now two trig equations that are linear in form to solve and get solutions for. Sine of theta equals 0.
00:26:29.959 --> 00:26:39.049
Now that's not going to produce, if we are looking for a reference angle, it's not going to produce pi over 3,
00:26:39.051 --> 00:26:49.392
pi over 4, or pi over 6, our special angles because if the ratio for the sine function is the opposite over the
00:26:49.394 --> 00:26:58.235
hypotenuse that will not form a triangle. So what I need to look at is the graph of my sine function.
00:26:58.237 --> 00:27:14.915
Let's graph from 0 to 2 pi, just a quick sketch. Then we will look and see where is the sine function 0?
00:27:14.917 --> 00:27:29.599
Sine function is 0, looking at the x-axis or the x-intercepts, is 0 at 0, pi, 2 pi, 3 pi, 4 pi, and continues,
00:27:29.601 --> 00:27:42.531
but we only want the solutions from 0 to 2 pi. Notice we do not want to include 2 pi, so for this equation,
00:27:42.533 --> 00:27:57.629
sine theta equals 0, we have two solutions at 0 and at pi. Ok, this equation now, we have the ratio
00:27:57.631 --> 00:28:06.626
negative 1/2, so we'll come to the side and find our reference angle. We're looking for pi over 3, pi over 4, or
00:28:06.628 --> 00:28:24.874
pi over 6. So we want to solve the equation the sine of theta r equals positive 1/2. Sine function, opposite 1,
00:28:24.876 --> 00:28:40.174
hypotenuse 2, adjacent side square root of 3, so the reference angle, pi over 6. That's right, but back to
00:28:40.176 --> 00:28:48.672
our problem. This ratio is negative, so we don't want an answer of pi over 6. We want the two quadrants
00:28:48.674 --> 00:29:01.357
where the sine function is negative. Quadrant three, that will give us a solution. Quadrant four, that will
00:29:01.359 --> 00:29:15.540
give us another solution. Quadrant three, we need pi plus the reference angle, pi plus pi over 6, which will
00:29:15.542 --> 00:29:28.129
give us 7 pi over 6, and then in the fourth quadrant, an answer for theta will be 2 pi minus the reference
00:29:28.131 --> 00:29:50.881
angle, which is 2 pi minus pi over 6, 11 pi over 6. Ok, this particular equation then has four solutions.
00:29:50.883 --> 00:30:06.339
Sometimes in writing your solutions you may be asked to put the solutions in ascending order, so just
00:30:06.341 --> 00:30:27.592
double check: 0, pi, then 7 pi over 6, then 11 pi over 6, and that is our final answer for this particular example.
00:30:30.000 --> 00:30:38.761
We're going to look at an example of a trigonometric equation that we will solve using identities. Our
00:30:38.763 --> 00:30:48.512
directions tell us to determine the specific solutions, if any, on the interval from 0 to 2 pi, and our example
00:30:48.514 --> 00:31:00.260
reads negative 3 times the sine of theta plus 3 equals 2 times the cosine of theta squared. The first thing
00:31:00.262 --> 00:31:10.260
we notice about this example is that we have the sine of theta and cosine of theta, two different trig functions.
00:31:10.262 --> 00:31:23.762
So what we need to do is find an identity that we can change either sines into cosines or cosines into sines.
00:31:23.764 --> 00:31:36.696
Let me think. Let me think. We have an identity that says sine squared theta plus cosine squared theta
00:31:36.698 --> 00:31:48.196
equals 1. Oh, that's a good one to use. We have the cosine of theta squared, so let's solve for the cosine
00:31:48.198 --> 00:32:02.946
of theta squared by subtracting the sine of theta squared from both sides. Then we can take this 1 minus
00:32:02.948 --> 00:32:20.150
sine of theta squared, and we can substitute it for the cosine of theta squared. I am going to put parentheses
00:32:20.152 --> 00:32:33.495
here to enter 1 minus the sine of theta squared because I'll need to come back and distribute the 2.
00:32:33.497 --> 00:32:53.271
Let's do that. Negative 3 sine theta plus 3 [equals] 2 times 1, 2 times negative sine of theta squared.
00:32:53.273 --> 00:33:02.519
Ok, let's see what we're going to do now. We need all of the terms on one side of the equation set equal
00:33:02.521 --> 00:33:13.203
to 0 so we can factor. The squared term, sine of theta squared, has a negative 2 in front, so I'm going to
00:33:13.205 --> 00:33:28.454
take that term to the left side of the equation by adding 2 sine of theta squared. Copy negative 3 sine of
00:33:28.456 --> 00:33:44.318
theta, and then subtract 2 from both sides. 3 minus 2 is plus 1. So we have a trigonometric equation that
00:33:44.320 --> 00:34:01.071
is in quadratic form, and we can factor. I notice the factors of 2 sine theta squared will be 2 sine theta and
00:34:01.073 --> 00:34:17.188
sine theta. Factors of 1 are 1 and 1, and then we need minus in both of those factors. Checking, this is
00:34:17.190 --> 00:34:28.026
negative sine theta. This is negative 2 sine theta, which is negative 3 sine theta. Now if you had to, you
00:34:28.028 --> 00:34:44.712
could come to the side and make a substitution for sine theta, and you would get the same factorization,
00:34:44.714 --> 00:34:54.963
but it might make it a little easier for you. Once we have our two factors set equal to 0, all we have to do,
00:34:54.965 --> 00:35:13.711
let's erase this, is solve each factor separately. 2 sine theta minus 1 equals 0. Set that factor equal to 0.
00:35:13.713 --> 00:35:34.962
Then set this factor, sine theta minus 1, equal to 0. Sine theta will then equal, add 1, divide by 2. Here, add 1.
00:35:34.964 --> 00:35:47.461
Ok, sine theta equals 1/2. We'll come to the side, find the reference angle, and then find the two quadrants
00:35:47.463 --> 00:36:03.537
in which the sine function is positive. Sine of the reference angle equals positive 1/2. Quadrant one,
00:36:03.539 --> 00:36:15.288
opposite side over hypotenuse. The adjacent side, then, is square root of 3, and the reference angle is
00:36:15.290 --> 00:36:29.537
pi over 6. Ok, that will give us one solution because this ratio is positive. Sine function is positive
00:36:29.539 --> 00:36:36.996
in quadrant one, same thing as the reference angle. In which other quadrant is the sine function positive?
00:36:36.998 --> 00:36:49.589
Well, that's quadrant two. That will give us a second solution. We find that angle by subtracting the
00:36:49.591 --> 00:37:11.587
reference angle from pi. Pi minus pi over 6 is 5 pi over 6. These are two solutions, solves this equation
00:37:11.589 --> 00:37:23.085
on this interval. Now we'll go to this part. Sine theta equals positive 1. We don't have a reference angle of
00:37:23.087 --> 00:37:37.587
pi over 3, pi over 4, pi over 6 for the sine of theta equals 1. What we need there is the graph of our sine
00:37:37.589 --> 00:38:00.337
function. So I'm going to graph from 0 to 2 pi. Well, let's do this a little better. There we go. That looks better.
00:38:00.339 --> 00:38:17.346
So, our range from negative 1 to 1, we notice that the sine of theta equals positive 1 at pi over 2. So
00:38:17.348 --> 00:38:27.437
summarizing, and we're going to write our final solution. On the interval from 0 to 2 pi, we have 3 answers,
00:38:27.439 --> 00:38:36.687
pi over 6, 5 pi over 6, pi over 2. I'm going to write the three solutions in ascending order, so I'm going to
00:38:36.689 --> 00:38:53.688
write pi over 6, pi over 2, and 5 pi over 6. Again, to solve this equation using identities, we made a
00:38:53.690 --> 00:39:00.900
substitution for cosine theta from our Pythagorean identity.
00:39:03.238 --> 00:39:13.437
This example involves us solving this trigonometric equation, cosine of 2 theta minus 3 equals negative 7
00:39:13.439 --> 00:39:23.686
times the cosine of theta. Our directions tell us to determine the specific solutions, if any, on the interval
00:39:23.688 --> 00:39:36.633
from 0 to 2 pi. We're going to use an identity because this trig function, cosine 2 theta, is different than this
00:39:36.635 --> 00:39:49.134
one which is the cosine of just theta. I want to use the identity for the cosine of 2 theta. There are actually
00:39:49.136 --> 00:40:03.385
three versions to this identity. We have cosine squared minus sine squared. We have 1 minus 2 sine of theta
00:40:03.387 --> 00:40:16.633
squared, and we have 2 cosine of theta squared minus 1. So I want to choose the one that will work the best
00:40:16.635 --> 00:40:27.632
for this problem. Since the equation has a cosine of theta function, we will use this version. I'm going to
00:40:27.634 --> 00:40:44.133
substitute 2 cosine of theta squared minus 1 for the cosine of 2 theta. Let's do that. 2 cosine of theta
00:40:44.135 --> 00:41:00.883
squared minus 1 minus 3 equals negative 7 cosine of theta. At this point, I need to combine like terms.
00:41:00.885 --> 00:41:10.484
I need to take the negative 7 cosine of theta to the left side, set the whole equation equal to 0.
00:41:10.486 --> 00:41:28.417
2 cosine of theta squared plus 7 cosine of theta, a negative 1, a negative 3, gives me negative 4. I have the terms
00:41:28.419 --> 00:41:45.418
in descending order, so I can factor. This factoring, I'm not sure what to do. Let me let u equal the cosine
00:41:45.420 --> 00:42:05.168
of theta. 2u squared plus 7u minus 4 equals 0. I know the factors of 2u squared are 2u and u, and I think
00:42:05.170 --> 00:42:18.167
the factors that I'm going to need for 4 are 4 and 1 with a plus sign here, minus sign there. Double check.
00:42:18.169 --> 00:42:31.669
On the outside I have 8u. On the inside I have negative 1u. That combines to positive 7u. Let's just make
00:42:31.671 --> 00:42:49.666
the substitution now with cosine theta. 2 cosine theta minus 1 times cosine theta plus 4. Once I have the
00:42:49.668 --> 00:43:06.208
product of these two factors set equal to 0, I can set each factor equal to 0 and solve.
00:43:06.210 --> 00:43:25.418
2 cosine theta minus 1 equals 0. I'm going to add 1 and divide by 2. Cosine theta plus 4, I simply subtract the 4.
00:43:25.420 --> 00:43:36.417
Cosine of theta equals 1/2. This is a positive ratio, so once I find the reference angle, I know that will be one
00:43:36.419 --> 00:43:46.383
solution to this equation on my given interval. But let's find the reference angle, so we can find the second
00:43:46.385 --> 00:44:03.633
solution. Cosine of theta r equals positive 1/2. Setting up our triangle, I know that the cosine function is the
00:44:03.635 --> 00:44:15.157
ratio of the adjacent side, I place the 1, divided by the hypotenuse, so I place the 2, and I know that the opposite
00:44:15.159 --> 00:44:29.909
side is the square root of 3. So the reference angle is pi over 3. Ok, where else is cosine function positive?
00:44:29.911 --> 00:44:45.409
Quadrant four. That angle we will find by subtracting the reference angle from 2 pi which gives us
00:44:45.411 --> 00:45:02.408
5 pi over 3. So these are, the pi over 3, the 5 pi over 3, are the first two solutions. Now, let's go to cosine
00:45:02.410 --> 00:45:17.959
theta equals negative 4. I know it's not pi over 3, pi over 4, pi over 6, so let me go to my graph of my
00:45:17.961 --> 00:45:39.709
cosine function, y equals cosine x. Maybe I will find solutions there. Here we go. From 0 to 2 pi.
00:45:39.711 --> 00:45:50.209
Oh, I see the problem now. The graph doesn't reach negative 4. We know that the range of this function
00:45:50.211 --> 00:46:02.959
is from negative 1 to 1. So this equation gives us no solutions on our given interval, or ever really.
00:46:02.961 --> 00:46:14.210
So we have two solutions, pi over 3, 5 pi over 3, on the interval from 0 to 2 pi.
00:46:15.900 --> 00:46:25.209
Now we want to solve a trigonometric equation using a calculator. Our directions are to approximate all
00:46:25.211 --> 00:46:32.959
solutions to the equation on the interval from 0 to 2 pi, and we're going to round our answers to four
00:46:32.961 --> 00:46:43.650
decimal places. Here is our example. Sine of theta is 78 hundredths. Here's the calculator. The first thing
00:46:43.652 --> 00:46:52.493
we need to know about the calculator is that we put it in radian mode. Make sure you know how to do that.
00:46:52.495 --> 00:47:10.429
Ok, got that. So let's write that in our notes: radian mode. This ratio is positive, so we want to first find
00:47:10.431 --> 00:47:30.358
the reference angle, and we're going to do that by entering in our calculator sine inverse of this positive ratio.
00:47:30.360 --> 00:47:47.858
So we're going to do second, sine, then .78, enter, and we get that the reference angle is approximately,
00:47:47.860 --> 00:48:03.974
let's change that, approximately .8947. Ok, I've rounded this 7. It was 6 6, and then I've rounded to 7.
00:48:03.976 --> 00:48:14.975
The ratio is positive so that will give us one solution, our reference angle. The second solution will be found
00:48:14.977 --> 00:48:28.727
in quadrant two, which we will find, let's call it theta two, by subtracting the reference angle from pi.
00:48:28.729 --> 00:49:01.723
So let's write our solutions: 0.8947, pi minus that number which according to our calculator is 2.2469.
00:49:01.725 --> 00:49:11.726
Those are the two solutions for the trig equation sine of theta equals 78 hundredths.