[Script Info]
Title:
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.97,0:00:05.57,Default,,0000,0000,0000,,In this video, we're going to be\Nsolving whole collection of
Dialogue: 0,0:00:05.57,0:00:09.33,Default,,0000,0000,0000,,trigonometric equations now be\Ncause it's the technique of
Dialogue: 0,0:00:09.33,0:00:13.51,Default,,0000,0000,0000,,solving the equation and in\Nensuring that we get enough
Dialogue: 0,0:00:13.51,0:00:17.27,Default,,0000,0000,0000,,solutions, that's important and\Nnot actually looking up the
Dialogue: 0,0:00:17.27,0:00:21.45,Default,,0000,0000,0000,,angle. All of these are designed\Naround certain special angles,
Dialogue: 0,0:00:21.45,0:00:26.89,Default,,0000,0000,0000,,so I'm just going to list at the\Nvery beginning here the special
Dialogue: 0,0:00:26.89,0:00:31.48,Default,,0000,0000,0000,,angles and their sines, cosines,\Nand tangents that are going to
Dialogue: 0,0:00:31.48,0:00:33.47,Default,,0000,0000,0000,,form. The basis of what\Nwe're doing.
Dialogue: 0,0:00:37.83,0:00:41.95,Default,,0000,0000,0000,,So the special angles that we're\Ngoing to have a look at our
Dialogue: 0,0:00:41.95,0:00:49.44,Default,,0000,0000,0000,,zero. 30 4560\Nand 90 there in degrees.
Dialogue: 0,0:00:49.44,0:00:55.24,Default,,0000,0000,0000,,If we're thinking about radians,\Nthen there's zero.
Dialogue: 0,0:00:55.94,0:00:59.35,Default,,0000,0000,0000,,Pie by 6.
Dialogue: 0,0:01:00.13,0:01:03.72,Default,,0000,0000,0000,,Pie by 4.
Dialogue: 0,0:01:04.67,0:01:05.74,Default,,0000,0000,0000,,Pie by three.
Dialogue: 0,0:01:06.37,0:01:09.09,Default,,0000,0000,0000,,And Π by 2.
Dialogue: 0,0:01:10.22,0:01:15.79,Default,,0000,0000,0000,,Trig ratios we're going to be\Nlooking at are the sign.
Dialogue: 0,0:01:15.79,0:01:21.10,Default,,0000,0000,0000,,The cosine. On the tangent\Nof each of these.
Dialogue: 0,0:01:22.46,0:01:25.85,Default,,0000,0000,0000,,Sign of 0 is 0.
Dialogue: 0,0:01:26.69,0:01:29.67,Default,,0000,0000,0000,,The sign of 30 is 1/2.
Dialogue: 0,0:01:30.37,0:01:34.34,Default,,0000,0000,0000,,Sign of 45 is one over Route 2.
Dialogue: 0,0:01:34.85,0:01:41.71,Default,,0000,0000,0000,,The sign of 60 is Route 3 over 2\Nand the sign of 90 is one.
Dialogue: 0,0:01:43.44,0:01:45.26,Default,,0000,0000,0000,,Cosine of 0 is one.
Dialogue: 0,0:01:45.99,0:01:52.66,Default,,0000,0000,0000,,Cosine of 30 is Route 3 over 2\Ncosine of 45 is one over Route
Dialogue: 0,0:01:52.66,0:01:58.90,Default,,0000,0000,0000,,2, the cosine of 60 is 1/2, and\Nthe cosine of 90 is 0.
Dialogue: 0,0:01:59.57,0:02:06.10,Default,,0000,0000,0000,,The town of 0 is 0 the\Ntown of 30 is one over
Dialogue: 0,0:02:06.10,0:02:12.62,Default,,0000,0000,0000,,Route 3 that Anna 45 is\N110 of 60 is Route 3 and
Dialogue: 0,0:02:12.62,0:02:17.14,Default,,0000,0000,0000,,the town of 90 degrees is\Ninfinite, it's undefined.
Dialogue: 0,0:02:18.63,0:02:21.74,Default,,0000,0000,0000,,It's these that we're going to\Nbe looking at and working with.
Dialogue: 0,0:02:22.56,0:02:25.90,Default,,0000,0000,0000,,Let's look at our first
Dialogue: 0,0:02:25.90,0:02:30.55,Default,,0000,0000,0000,,equation then. We're going to\Nbegin with some very simple
Dialogue: 0,0:02:30.55,0:02:37.84,Default,,0000,0000,0000,,ones. So we take sign of\NX is equal to nought .5. Now
Dialogue: 0,0:02:37.84,0:02:43.75,Default,,0000,0000,0000,,invariably when we get an\Nequation we get a range of
Dialogue: 0,0:02:43.75,0:02:45.90,Default,,0000,0000,0000,,values along with it.
Dialogue: 0,0:02:46.61,0:02:52.71,Default,,0000,0000,0000,,So in this case will take X is\Nbetween North and 360. So what
Dialogue: 0,0:02:52.71,0:02:56.64,Default,,0000,0000,0000,,we're looking for is all the\Nvalues of X.
Dialogue: 0,0:02:57.20,0:03:01.46,Default,,0000,0000,0000,,Husain gives us N\N.5.
Dialogue: 0,0:03:03.97,0:03:10.83,Default,,0000,0000,0000,,Let's sketch a graph of\Nsine X over this range.
Dialogue: 0,0:03:13.61,0:03:16.16,Default,,0000,0000,0000,,And sign looks like that\Nwith 90.
Dialogue: 0,0:03:17.31,0:03:18.50,Default,,0000,0000,0000,,180
Dialogue: 0,0:03:20.12,0:03:27.15,Default,,0000,0000,0000,,270 and 360 and ranging between\None 4 sign 90 and minus
Dialogue: 0,0:03:27.15,0:03:30.67,Default,,0000,0000,0000,,one for the sign of 270.
Dialogue: 0,0:03:31.36,0:03:36.61,Default,,0000,0000,0000,,Sign of X is nought .5. So\Nwe go there.
Dialogue: 0,0:03:37.89,0:03:38.72,Default,,0000,0000,0000,,And there.
Dialogue: 0,0:03:39.78,0:03:44.41,Default,,0000,0000,0000,,So there's our first angle, and\Nthere's our second angle.
Dialogue: 0,0:03:45.90,0:03:52.52,Default,,0000,0000,0000,,We know the first one is 30\Ndegrees because sign of 30 is
Dialogue: 0,0:03:52.52,0:03:58.62,Default,,0000,0000,0000,,1/2, so our first angle is 30\Ndegrees. This curve is symmetric
Dialogue: 0,0:03:58.62,0:04:05.24,Default,,0000,0000,0000,,and so because were 30 degrees\Nin from there, this one's got to
Dialogue: 0,0:04:05.24,0:04:08.30,Default,,0000,0000,0000,,be 30 degrees back from there.
Dialogue: 0,0:04:08.81,0:04:12.61,Default,,0000,0000,0000,,That would make it
Dialogue: 0,0:04:12.61,0:04:17.08,Default,,0000,0000,0000,,150. There are no more answers\Nbecause within this range as we
Dialogue: 0,0:04:17.08,0:04:18.36,Default,,0000,0000,0000,,go along this line.
Dialogue: 0,0:04:18.89,0:04:23.45,Default,,0000,0000,0000,,It doesn't cross the curve at\Nany other points.
Dialogue: 0,0:04:23.46,0:04:30.00,Default,,0000,0000,0000,,Let's have a look\Nat a cosine cause
Dialogue: 0,0:04:30.00,0:04:36.55,Default,,0000,0000,0000,,of X is minus\Nnought .5 and the
Dialogue: 0,0:04:36.55,0:04:43.09,Default,,0000,0000,0000,,range for this X\Nbetween North and 360.
Dialogue: 0,0:04:43.82,0:04:49.11,Default,,0000,0000,0000,,So again, let's have a look at a\Ngraph of the function.
Dialogue: 0,0:04:50.07,0:04:53.90,Default,,0000,0000,0000,,Involved in the equation,\Nthe cosine graph.
Dialogue: 0,0:04:56.51,0:05:00.09,Default,,0000,0000,0000,,Looks like that. One and
Dialogue: 0,0:05:00.09,0:05:04.82,Default,,0000,0000,0000,,minus one.\NThis is 90.
Dialogue: 0,0:05:06.30,0:05:08.16,Default,,0000,0000,0000,,180
Dialogue: 0,0:05:09.31,0:05:15.84,Default,,0000,0000,0000,,270 and then here\Nat the end, 360.
Dialogue: 0,0:05:17.02,0:05:19.77,Default,,0000,0000,0000,,Minus 9.5.
Dialogue: 0,0:05:20.81,0:05:26.39,Default,,0000,0000,0000,,Gain across there at minus\N9.5 and down to their and
Dialogue: 0,0:05:26.39,0:05:27.91,Default,,0000,0000,0000,,down to their.
Dialogue: 0,0:05:28.97,0:05:35.45,Default,,0000,0000,0000,,Now the one thing we do know is\Nthat the cause of 60 is plus N
Dialogue: 0,0:05:35.45,0:05:40.72,Default,,0000,0000,0000,,.5, and so that's there. So we\Nknow there is 60. Now again,
Dialogue: 0,0:05:40.72,0:05:45.98,Default,,0000,0000,0000,,this curve is symmetric, so if\Nthat one is 30 back that way
Dialogue: 0,0:05:45.98,0:05:51.65,Default,,0000,0000,0000,,this one must be 30 further on.\NSo I'll first angle must be 120
Dialogue: 0,0:05:51.65,0:05:56.92,Default,,0000,0000,0000,,degrees. This one's got to be in\Na similar position as this bit
Dialogue: 0,0:05:56.92,0:05:59.34,Default,,0000,0000,0000,,of the curve is again symmetric.
Dialogue: 0,0:05:59.38,0:06:07.22,Default,,0000,0000,0000,,So that's 270 and we need\Nto come back 30 degrees, so
Dialogue: 0,0:06:07.22,0:06:12.51,Default,,0000,0000,0000,,that's 240. Now we're going to\Nhave a look at an example where
Dialogue: 0,0:06:12.51,0:06:16.49,Default,,0000,0000,0000,,we've got what we call on\Nmultiple angle. So instead of
Dialogue: 0,0:06:16.49,0:06:21.56,Default,,0000,0000,0000,,just being cause of X or sign of\NX, it's going to be something
Dialogue: 0,0:06:21.56,0:06:26.27,Default,,0000,0000,0000,,like sign of 2X or cause of\Nthree X. So let's begin with
Dialogue: 0,0:06:26.27,0:06:33.08,Default,,0000,0000,0000,,sign of. 2X is equal\Nto Route 3 over 2
Dialogue: 0,0:06:33.08,0:06:40.05,Default,,0000,0000,0000,,and again will take X\Nto be between North and
Dialogue: 0,0:06:40.05,0:06:40.74,Default,,0000,0000,0000,,360.
Dialogue: 0,0:06:41.81,0:06:44.84,Default,,0000,0000,0000,,Now we've got 2X here.
Dialogue: 0,0:06:45.52,0:06:52.50,Default,,0000,0000,0000,,So if we've got 2X and X\Nis between Norton 360, then the
Dialogue: 0,0:06:52.50,0:06:59.48,Default,,0000,0000,0000,,total range that we're going to\Nbe looking at is not to 722.
Dialogue: 0,0:06:59.48,0:07:06.46,Default,,0000,0000,0000,,X is going to come between 0\Nand 720, and the sign function
Dialogue: 0,0:07:06.46,0:07:12.37,Default,,0000,0000,0000,,is periodic. It repeats itself\Nevery 360 degrees, so I'm going
Dialogue: 0,0:07:12.37,0:07:16.13,Default,,0000,0000,0000,,to need 2 copies of the sine
Dialogue: 0,0:07:16.13,0:07:22.41,Default,,0000,0000,0000,,curve. As the first one going up\Nto 360 and now I need a second
Dialogue: 0,0:07:22.41,0:07:24.46,Default,,0000,0000,0000,,copy there going on till.
Dialogue: 0,0:07:25.24,0:07:27.32,Default,,0000,0000,0000,,720
Dialogue: 0,0:07:28.37,0:07:34.54,Default,,0000,0000,0000,,OK, so sign 2 X equals root, 3\Nover 2, but we know that the
Dialogue: 0,0:07:34.54,0:07:41.93,Default,,0000,0000,0000,,sign of 60 is Route 3 over 2. So\Nif we put in Route 3 over 2 it's
Dialogue: 0,0:07:41.93,0:07:46.86,Default,,0000,0000,0000,,there, then it's going to be\Nthese along here as well. So
Dialogue: 0,0:07:46.86,0:07:52.21,Default,,0000,0000,0000,,what have we got? Well, the\Nfirst one here we know is 60.
Dialogue: 0,0:07:52.21,0:07:57.96,Default,,0000,0000,0000,,This point we know is 180 so\Nthat one's got to be the same
Dialogue: 0,0:07:57.96,0:08:04.18,Default,,0000,0000,0000,,distance. Back in due to the\Nsymmetry 120, so we do know that
Dialogue: 0,0:08:04.18,0:08:10.93,Default,,0000,0000,0000,,2X will be 60 or 120, but we\Nalso now we've got these other
Dialogue: 0,0:08:10.93,0:08:16.72,Default,,0000,0000,0000,,points on here, so let's just\Ncount on where we are. There's
Dialogue: 0,0:08:16.72,0:08:22.02,Default,,0000,0000,0000,,the 1st loop of the sign\Nfunction, the first copy, its
Dialogue: 0,0:08:22.02,0:08:27.32,Default,,0000,0000,0000,,periodic and repeats itself\Nagain. So now we need to know
Dialogue: 0,0:08:27.32,0:08:29.25,Default,,0000,0000,0000,,where are these well.
Dialogue: 0,0:08:29.27,0:08:36.30,Default,,0000,0000,0000,,This is an exact copy of\Nthat, so this must be 60
Dialogue: 0,0:08:36.30,0:08:43.33,Default,,0000,0000,0000,,further on. In other words, at\N420, and this must be another
Dialogue: 0,0:08:43.33,0:08:50.37,Default,,0000,0000,0000,,120 further on. In other words,\Nat 480. So we've got two
Dialogue: 0,0:08:50.37,0:08:57.21,Default,,0000,0000,0000,,more answers. And it's X\Nthat we actually want, not
Dialogue: 0,0:08:57.21,0:09:00.64,Default,,0000,0000,0000,,2X. So this is 3060.
Dialogue: 0,0:09:00.82,0:09:04.44,Default,,0000,0000,0000,,210 and finally
Dialogue: 0,0:09:04.44,0:09:11.66,Default,,0000,0000,0000,,240. Let's have a\Nlook at that with a tangent
Dialogue: 0,0:09:11.66,0:09:17.59,Default,,0000,0000,0000,,function. This time tan or three\NX is equal to.
Dialogue: 0,0:09:18.16,0:09:24.86,Default,,0000,0000,0000,,Minus one and will\Ntake X to be
Dialogue: 0,0:09:24.86,0:09:28.22,Default,,0000,0000,0000,,between North and 180.
Dialogue: 0,0:09:29.74,0:09:36.04,Default,,0000,0000,0000,,So we draw a graph\Nof the tangent function.
Dialogue: 0,0:09:37.14,0:09:38.38,Default,,0000,0000,0000,,So we go up.
Dialogue: 0,0:09:40.99,0:09:43.78,Default,,0000,0000,0000,,We've got that there. That's 90.
Dialogue: 0,0:09:53.09,0:09:59.84,Default,,0000,0000,0000,,This is 180 and this is 270\Nnow. It's 3X. X is between
Dialogue: 0,0:09:59.84,0:10:07.10,Default,,0000,0000,0000,,Norton 180, so 3X can be between\NNorth and 3 * 180 which is
Dialogue: 0,0:10:07.10,0:10:13.85,Default,,0000,0000,0000,,540. So I need to get copies\Nof this using the periodicity of
Dialogue: 0,0:10:13.85,0:10:20.60,Default,,0000,0000,0000,,the tangent function right up to\N540. So let's put in some more.
Dialogue: 0,0:10:21.81,0:10:26.61,Default,,0000,0000,0000,,That's 360. On\Nthere.
Dialogue: 0,0:10:27.92,0:10:30.90,Default,,0000,0000,0000,,That's 450.
Dialogue: 0,0:10:34.82,0:10:42.47,Default,,0000,0000,0000,,This one here will be\N540 and that's as near
Dialogue: 0,0:10:42.47,0:10:50.12,Default,,0000,0000,0000,,or as far as we\Nneed to go. Tanner 3X
Dialogue: 0,0:10:50.12,0:10:53.94,Default,,0000,0000,0000,,is minus one, so here's
Dialogue: 0,0:10:53.94,0:10:59.12,Default,,0000,0000,0000,,minus one. And we go across here\Npicking off all the ones that we
Dialogue: 0,0:10:59.12,0:11:00.94,Default,,0000,0000,0000,,need. So we've got one there.
Dialogue: 0,0:11:01.69,0:11:08.86,Default,,0000,0000,0000,,There there. These are our\Nvalues, so 3X is equal 12.
Dialogue: 0,0:11:08.86,0:11:16.69,Default,,0000,0000,0000,,Now we know that the angle\Nwhose tangent is one is 45,
Dialogue: 0,0:11:16.69,0:11:24.51,Default,,0000,0000,0000,,which is there. So again this\Nand this are symmetric bits of
Dialogue: 0,0:11:24.51,0:11:32.33,Default,,0000,0000,0000,,curve, so this must be 45\Nfurther on. In other words 130.
Dialogue: 0,0:11:32.34,0:11:32.98,Default,,0000,0000,0000,,5.
Dialogue: 0,0:11:34.17,0:11:41.95,Default,,0000,0000,0000,,This one here has got to be\N45 further on, so that will be
Dialogue: 0,0:11:41.95,0:11:49.67,Default,,0000,0000,0000,,315. This one here has got\Nto be 45 further on, so that
Dialogue: 0,0:11:49.67,0:11:57.44,Default,,0000,0000,0000,,will be 495, but it's X that\Nwe want not 3X, so let's divide
Dialogue: 0,0:11:57.44,0:12:04.10,Default,,0000,0000,0000,,throughout by three, so freezing\Nto that is 45 threes into that
Dialogue: 0,0:12:04.10,0:12:11.32,Default,,0000,0000,0000,,is 105 and threes into that is\N165. Those are our three answers
Dialogue: 0,0:12:11.32,0:12:14.09,Default,,0000,0000,0000,,for that one 45 degrees.
Dialogue: 0,0:12:14.10,0:12:17.11,Default,,0000,0000,0000,,105 degrees under
Dialogue: 0,0:12:17.11,0:12:24.68,Default,,0000,0000,0000,,165. Let's take cause of\NX over 2 this time. So
Dialogue: 0,0:12:24.68,0:12:31.96,Default,,0000,0000,0000,,instead of multiplying by two or\Nby three, were now dividing by
Dialogue: 0,0:12:31.96,0:12:38.64,Default,,0000,0000,0000,,two. Let's see what difference\Nthis might make equals minus 1/2
Dialogue: 0,0:12:38.64,0:12:45.92,Default,,0000,0000,0000,,and will take X to be\Nbetween North and 360. So let's
Dialogue: 0,0:12:45.92,0:12:47.75,Default,,0000,0000,0000,,draw the graph.
Dialogue: 0,0:12:48.83,0:12:55.97,Default,,0000,0000,0000,,All calls X between North\Nand 360, so there we've
Dialogue: 0,0:12:55.97,0:12:58.83,Default,,0000,0000,0000,,got it 360 there.
Dialogue: 0,0:12:59.56,0:13:06.35,Default,,0000,0000,0000,,180 there, we've got 90 and 270\Nthere in their minus. 1/2 now
Dialogue: 0,0:13:06.35,0:13:08.43,Default,,0000,0000,0000,,that's going to be.
Dialogue: 0,0:13:09.99,0:13:17.47,Default,,0000,0000,0000,,Their cross and then these are\Nthe ones that we are after.
Dialogue: 0,0:13:19.06,0:13:25.68,Default,,0000,0000,0000,,So let's work with that. X over\N2 is equal tool. Now where are
Dialogue: 0,0:13:25.68,0:13:32.30,Default,,0000,0000,0000,,we? Well, we know that the angle\Nwhose cosine is 1/2 is in fact
Dialogue: 0,0:13:32.30,0:13:38.93,Default,,0000,0000,0000,,60 degrees, which is here 30 in\Nfrom there. So that must be 30
Dialogue: 0,0:13:38.93,0:13:45.55,Default,,0000,0000,0000,,further on. In other words, 120\Nand this one must be 30 back. In
Dialogue: 0,0:13:45.55,0:13:49.80,Default,,0000,0000,0000,,other words, 240. So now we\Nmultiply it by.
Dialogue: 0,0:13:49.85,0:13:57.08,Default,,0000,0000,0000,,Two, we get 240 and 480, but\Nof course this one is outside
Dialogue: 0,0:13:57.08,0:14:04.31,Default,,0000,0000,0000,,the given range. The range is\Nnot to 360, so we do not
Dialogue: 0,0:14:04.31,0:14:07.64,Default,,0000,0000,0000,,need that answer, just want the
Dialogue: 0,0:14:07.64,0:14:14.70,Default,,0000,0000,0000,,240. Now we've been working with\Na range of North 360, or in one
Dialogue: 0,0:14:14.70,0:14:20.48,Default,,0000,0000,0000,,case not to 180, so let's change\Nthe range now so it's a
Dialogue: 0,0:14:20.48,0:14:26.71,Default,,0000,0000,0000,,symmetric range in the Y axis,\Nso the range is now going to run
Dialogue: 0,0:14:26.71,0:14:29.38,Default,,0000,0000,0000,,from minus 180 to plus 180
Dialogue: 0,0:14:29.38,0:14:36.47,Default,,0000,0000,0000,,degrees. So we'll begin with\Nsign of X equals 1X is to
Dialogue: 0,0:14:36.47,0:14:41.89,Default,,0000,0000,0000,,be between 180 degrees but\Ngreater than minus 180 degrees.
Dialogue: 0,0:14:41.89,0:14:48.94,Default,,0000,0000,0000,,Let's sketch the graph of sign\Nin that range. So we want to
Dialogue: 0,0:14:48.94,0:14:54.36,Default,,0000,0000,0000,,complete copy of it. It's going\Nto look like that.
Dialogue: 0,0:14:55.20,0:15:00.19,Default,,0000,0000,0000,,Now we know that the angle\Nwho sign is one is 90
Dialogue: 0,0:15:00.19,0:15:04.35,Default,,0000,0000,0000,,degrees and so we know\Nthat's one there and that's
Dialogue: 0,0:15:04.35,0:15:09.34,Default,,0000,0000,0000,,90 there and we can see that\Nthere is only the one
Dialogue: 0,0:15:09.34,0:15:13.50,Default,,0000,0000,0000,,solution it meets the curve\Nonce and once only, so
Dialogue: 0,0:15:13.50,0:15:14.75,Default,,0000,0000,0000,,that's 90 degrees.
Dialogue: 0,0:15:15.92,0:15:22.33,Default,,0000,0000,0000,,Once and once only, that is\Nwithin the defined range. Let's
Dialogue: 0,0:15:22.33,0:15:24.08,Default,,0000,0000,0000,,take another one.
Dialogue: 0,0:15:24.23,0:15:31.63,Default,,0000,0000,0000,,So now we use a multiple\Nangle cause 2 X equals 1/2
Dialogue: 0,0:15:31.63,0:15:38.42,Default,,0000,0000,0000,,and will take X to be\Nbetween minus 180 degrees and
Dialogue: 0,0:15:38.42,0:15:44.59,Default,,0000,0000,0000,,plus 180 degrees. So let's\Nsketch the graph. Let's remember
Dialogue: 0,0:15:44.59,0:15:51.100,Default,,0000,0000,0000,,that if X is between minus\N180 and plus one 80, then
Dialogue: 0,0:15:51.100,0:15:55.70,Default,,0000,0000,0000,,2X will be between minus 360.
Dialogue: 0,0:15:55.74,0:15:57.93,Default,,0000,0000,0000,,And plus 360.
Dialogue: 0,0:16:02.83,0:16:07.84,Default,,0000,0000,0000,,So what we need to do is use the\Nperiodicity of the cosine
Dialogue: 0,0:16:07.84,0:16:09.38,Default,,0000,0000,0000,,function to sketch it.
Dialogue: 0,0:16:09.98,0:16:14.63,Default,,0000,0000,0000,,In the range. So there's\Nthe knocked 360 bit and
Dialogue: 0,0:16:14.63,0:16:16.02,Default,,0000,0000,0000,,then we want.
Dialogue: 0,0:16:19.55,0:16:25.97,Default,,0000,0000,0000,,To minus 360. So I just label up\Nthe points. Here is 90.
Dialogue: 0,0:16:26.94,0:16:28.09,Default,,0000,0000,0000,,180
Dialogue: 0,0:16:29.24,0:16:36.54,Default,,0000,0000,0000,,Two 7360 and then back\Nthis way minus 90 -
Dialogue: 0,0:16:36.54,0:16:40.34,Default,,0000,0000,0000,,180. Minus 270 and
Dialogue: 0,0:16:40.34,0:16:47.63,Default,,0000,0000,0000,,minus 360. Now cause\N2X is 1/2, so here's a half.
Dialogue: 0,0:16:48.26,0:16:52.74,Default,,0000,0000,0000,,Membrane that this goes between\Nplus one and minus one and if we
Dialogue: 0,0:16:52.74,0:16:56.54,Default,,0000,0000,0000,,draw a line across to see where\Nit meets the curve.
Dialogue: 0,0:16:58.53,0:17:04.90,Default,,0000,0000,0000,,Then we can see it meets it in\Nfour places. There, there there
Dialogue: 0,0:17:04.90,0:17:11.27,Default,,0000,0000,0000,,and there we know that the angle\Nwhere it meets here is 60
Dialogue: 0,0:17:11.27,0:17:16.66,Default,,0000,0000,0000,,degrees. So our first value is 2\NX equals 60 degrees.
Dialogue: 0,0:17:17.60,0:17:23.77,Default,,0000,0000,0000,,By symmetry, this one back here\Nhas got to be minus 60.
Dialogue: 0,0:17:24.35,0:17:30.49,Default,,0000,0000,0000,,What about this one here? Well,\Nagain, symmetry says that we are
Dialogue: 0,0:17:30.49,0:17:37.66,Default,,0000,0000,0000,,60 from here, so we've got to\Nbe 60 back from there, so this
Dialogue: 0,0:17:37.66,0:17:44.32,Default,,0000,0000,0000,,must be 300 and our symmetry of\Nthe curve says that this one
Dialogue: 0,0:17:44.32,0:17:51.49,Default,,0000,0000,0000,,must be minus 300, and so we\Nhave X is 30 degrees minus 30
Dialogue: 0,0:17:51.49,0:17:54.05,Default,,0000,0000,0000,,degrees, 150 degrees and minus
Dialogue: 0,0:17:54.05,0:18:00.82,Default,,0000,0000,0000,,150 degrees. Working with\Nthe tangent function tan, two
Dialogue: 0,0:18:00.82,0:18:07.74,Default,,0000,0000,0000,,X equals Route 3 and\Nagain will place X between
Dialogue: 0,0:18:07.74,0:18:14.66,Default,,0000,0000,0000,,180 degrees and minus 180\Ndegrees. We want to sketch
Dialogue: 0,0:18:14.66,0:18:21.58,Default,,0000,0000,0000,,the function for tangent and\Nwe want to be aware
Dialogue: 0,0:18:21.58,0:18:24.35,Default,,0000,0000,0000,,that we've got 2X.
Dialogue: 0,0:18:24.97,0:18:32.24,Default,,0000,0000,0000,,So since X is between minus 118\N+ 182, X is got to be between
Dialogue: 0,0:18:32.24,0:18:34.67,Default,,0000,0000,0000,,minus 360 and plus 360.
Dialogue: 0,0:18:38.67,0:18:40.76,Default,,0000,0000,0000,,So if we take the bit between.
Dialogue: 0,0:18:45.86,0:18:48.63,Default,,0000,0000,0000,,North And 360.
Dialogue: 0,0:18:49.53,0:18:55.13,Default,,0000,0000,0000,,Which is that bit of the curve\Nwe need a copy of that between
Dialogue: 0,0:18:55.13,0:18:59.13,Default,,0000,0000,0000,,minus 360 and 0 because again\Nthe tangent function is
Dialogue: 0,0:18:59.13,0:19:01.53,Default,,0000,0000,0000,,periodic, so we need this bit.
Dialogue: 0,0:19:07.58,0:19:08.51,Default,,0000,0000,0000,,That
Dialogue: 0,0:19:14.62,0:19:18.68,Default,,0000,0000,0000,,And we need that and it's Mark\Noff this axis so we know where
Dialogue: 0,0:19:18.68,0:19:20.13,Default,,0000,0000,0000,,we are. This is 90.
Dialogue: 0,0:19:21.61,0:19:23.05,Default,,0000,0000,0000,,180
Dialogue: 0,0:19:24.28,0:19:31.79,Default,,0000,0000,0000,,270 and 360. So this\Nmust be minus 90 -
Dialogue: 0,0:19:31.79,0:19:36.30,Default,,0000,0000,0000,,180 - 270 and minus\N360.
Dialogue: 0,0:19:37.42,0:19:42.67,Default,,0000,0000,0000,,Now 2X is Route 3, the angle\Nwhose tangent is Route 3. We
Dialogue: 0,0:19:42.67,0:19:49.14,Default,,0000,0000,0000,,know is 60, so we go across here\Nat Route 3 and we meet the curve
Dialogue: 0,0:19:49.14,0:19:50.35,Default,,0000,0000,0000,,there and there.
Dialogue: 0,0:19:51.44,0:19:57.66,Default,,0000,0000,0000,,And we come back this way. We\Nmeet it there and we meet there.
Dialogue: 0,0:19:57.66,0:20:00.32,Default,,0000,0000,0000,,So our answers are down here.
Dialogue: 0,0:20:01.09,0:20:07.59,Default,,0000,0000,0000,,Working with this one, first we\Nknow that that is 60, so 2X is
Dialogue: 0,0:20:07.59,0:20:14.08,Default,,0000,0000,0000,,equal to 60 and so that that one\Nis 60 degrees on from that
Dialogue: 0,0:20:14.08,0:20:19.65,Default,,0000,0000,0000,,point. Symmetry says there for\Nthis one is also 60 degrees on
Dialogue: 0,0:20:19.65,0:20:22.43,Default,,0000,0000,0000,,from there. In other words, it's
Dialogue: 0,0:20:22.43,0:20:28.69,Default,,0000,0000,0000,,240. Let's work our way\Nbackwards. This one must be 60
Dialogue: 0,0:20:28.69,0:20:36.05,Default,,0000,0000,0000,,degrees on from minus 180, so it\Nmust be at minus 120. This one
Dialogue: 0,0:20:36.05,0:20:38.16,Default,,0000,0000,0000,,is 60 degrees on.
Dialogue: 0,0:20:39.22,0:20:46.42,Default,,0000,0000,0000,,From minus 360 and so therefore\Nit must be minus 300.
Dialogue: 0,0:20:47.12,0:20:54.19,Default,,0000,0000,0000,,And so if we divide throughout\Nby two, we have 31120 -
Dialogue: 0,0:20:54.19,0:21:01.26,Default,,0000,0000,0000,,60 and minus 150 degrees. We\Nwant to put degree signs on
Dialogue: 0,0:21:01.26,0:21:06.56,Default,,0000,0000,0000,,all of these, so there are\Nfour solutions there.
Dialogue: 0,0:21:07.47,0:21:12.86,Default,,0000,0000,0000,,Trick equations often come up as\Na result of having expressions
Dialogue: 0,0:21:12.86,0:21:17.27,Default,,0000,0000,0000,,or other equations which are\Nrather more complicated than
Dialogue: 0,0:21:17.27,0:21:19.23,Default,,0000,0000,0000,,that and depends upon
Dialogue: 0,0:21:19.23,0:21:26.36,Default,,0000,0000,0000,,identity's. So I'm going to\Nhave a look at a couple
Dialogue: 0,0:21:26.36,0:21:31.80,Default,,0000,0000,0000,,of equations. These equations\Nboth dependa pawn two identity's
Dialogue: 0,0:21:31.80,0:21:37.86,Default,,0000,0000,0000,,that is expressions involving\Ntrig functions that are true for
Dialogue: 0,0:21:37.86,0:21:40.28,Default,,0000,0000,0000,,all values of X.
Dialogue: 0,0:21:40.84,0:21:45.81,Default,,0000,0000,0000,,So the first one is sine\Nsquared of X plus cost
Dialogue: 0,0:21:45.81,0:21:51.69,Default,,0000,0000,0000,,squared of X is one. This is\Ntrue for all values of X.
Dialogue: 0,0:21:52.89,0:21:56.13,Default,,0000,0000,0000,,The second one we derive from
Dialogue: 0,0:21:56.13,0:22:02.12,Default,,0000,0000,0000,,this one. How we derive it\Ndoesn't matter at the moment,
Dialogue: 0,0:22:02.12,0:22:09.18,Default,,0000,0000,0000,,but what it tells us is that sex\Nsquared X is equal to 1 + 10
Dialogue: 0,0:22:09.18,0:22:15.34,Default,,0000,0000,0000,,squared X. So these are the two\Nidentity's that I'm going to be
Dialogue: 0,0:22:15.34,0:22:21.97,Default,,0000,0000,0000,,using. Sine squared X plus cost\Nsquared X is one and sex squared
Dialogue: 0,0:22:21.97,0:22:25.55,Default,,0000,0000,0000,,of X is 1 + 10 squared of
Dialogue: 0,0:22:25.55,0:22:33.07,Default,,0000,0000,0000,,X OK. So how do we go\Nabout using one of those to do
Dialogue: 0,0:22:33.07,0:22:35.77,Default,,0000,0000,0000,,an equation like this? Cos
Dialogue: 0,0:22:35.77,0:22:42.46,Default,,0000,0000,0000,,squared X? Plus cause\Nof X is equal
Dialogue: 0,0:22:42.46,0:22:49.08,Default,,0000,0000,0000,,to sine squared of\NX&X is between 180
Dialogue: 0,0:22:49.08,0:22:51.56,Default,,0000,0000,0000,,and 0 degrees.
Dialogue: 0,0:22:52.69,0:22:53.31,Default,,0000,0000,0000,,Well.
Dialogue: 0,0:22:54.75,0:22:57.86,Default,,0000,0000,0000,,We've got a cost squared, A\Ncause and a sine squared.
Dialogue: 0,0:22:58.68,0:23:03.85,Default,,0000,0000,0000,,If we were to use our identity\Nsine squared plus cost squared
Dialogue: 0,0:23:03.85,0:23:08.59,Default,,0000,0000,0000,,is one to replace the sine\Nsquared. Here I'd have a
Dialogue: 0,0:23:08.59,0:23:15.06,Default,,0000,0000,0000,,quadratic in terms of Cos X, and\Nif I got a quadratic then I know
Dialogue: 0,0:23:15.06,0:23:19.80,Default,,0000,0000,0000,,I can solve it either by\NFactorizing or by using the
Dialogue: 0,0:23:19.80,0:23:24.54,Default,,0000,0000,0000,,formula. So let me write down\Nsign squared X plus cost
Dialogue: 0,0:23:24.54,0:23:29.28,Default,,0000,0000,0000,,squared. X is equal to 1, from\Nwhich we can see.
Dialogue: 0,0:23:29.31,0:23:36.52,Default,,0000,0000,0000,,Sine squared X is equal to\N1 minus Cos squared of X,
Dialogue: 0,0:23:36.52,0:23:43.73,Default,,0000,0000,0000,,so I can take this and\Nplug it into their. So my
Dialogue: 0,0:23:43.73,0:23:49.74,Default,,0000,0000,0000,,equation now becomes cost\Nsquared X Plus X is equal
Dialogue: 0,0:23:49.74,0:23:53.35,Default,,0000,0000,0000,,to 1 minus Cos squared X.
Dialogue: 0,0:23:54.09,0:24:00.72,Default,,0000,0000,0000,,I want to get this as\Na quadratic square term linear
Dialogue: 0,0:24:00.72,0:24:07.96,Default,,0000,0000,0000,,term. Constant term equals 0, so\NI begin by adding cost squared
Dialogue: 0,0:24:07.96,0:24:09.77,Default,,0000,0000,0000,,to both sides.
Dialogue: 0,0:24:09.87,0:24:16.33,Default,,0000,0000,0000,,So adding on a cost squared\Nthere makes 2 Cos squared X plus
Dialogue: 0,0:24:16.33,0:24:23.29,Default,,0000,0000,0000,,cause X equals 1. 'cause I added\Ncost square to get rid of that
Dialogue: 0,0:24:23.29,0:24:30.25,Default,,0000,0000,0000,,one. Now I need to take one away\Nfrom both sides to cost squared
Dialogue: 0,0:24:30.25,0:24:33.73,Default,,0000,0000,0000,,X Plus X minus one equals 0.
Dialogue: 0,0:24:34.85,0:24:38.96,Default,,0000,0000,0000,,Now this is just a quadratic\Nequation, so the first question
Dialogue: 0,0:24:38.96,0:24:43.83,Default,,0000,0000,0000,,I've got to ask is does it\Nfactorize? So let's see if we
Dialogue: 0,0:24:43.83,0:24:45.70,Default,,0000,0000,0000,,can get it to factorize.
Dialogue: 0,0:24:46.55,0:24:51.30,Default,,0000,0000,0000,,I'll put two calls X in there\Nand cause X in there because
Dialogue: 0,0:24:51.30,0:24:56.77,Default,,0000,0000,0000,,that 2 cause X times that cause\NX gives Me 2 cost squared and I
Dialogue: 0,0:24:56.77,0:25:01.88,Default,,0000,0000,0000,,put a one under one there 'cause\None times by one gives me one
Dialogue: 0,0:25:01.88,0:25:07.36,Default,,0000,0000,0000,,and now I know to get a minus\Nsign. One's got to be minus and
Dialogue: 0,0:25:07.36,0:25:13.20,Default,,0000,0000,0000,,one's got to be plus now I want\Nplus cause X so if I make this
Dialogue: 0,0:25:13.20,0:25:16.84,Default,,0000,0000,0000,,one plus I'll have two cause X\Ntimes by one.
Dialogue: 0,0:25:16.88,0:25:21.66,Default,,0000,0000,0000,,Is to cause X if I make this one\Nminus I'll have minus Cos X from
Dialogue: 0,0:25:21.66,0:25:26.33,Default,,0000,0000,0000,,there. Taking those two\Ntogether, +2 cause X minus Cos X
Dialogue: 0,0:25:26.33,0:25:31.30,Default,,0000,0000,0000,,is going to give me the plus\NKozaks in there, so that equals
Dialogue: 0,0:25:31.30,0:25:36.29,Default,,0000,0000,0000,,0. Now, if not equal 0, I'm\Nmultiplying 2 numbers together.
Dialogue: 0,0:25:36.29,0:25:42.29,Default,,0000,0000,0000,,This one 2 cause X minus one and\Nthis one cause X plus one, so
Dialogue: 0,0:25:42.29,0:25:47.89,Default,,0000,0000,0000,,one of them or both of them have\Ngot to be equal to 0.
Dialogue: 0,0:25:48.77,0:25:54.71,Default,,0000,0000,0000,,So 2 calls X minus\None is 0.
Dialogue: 0,0:25:55.56,0:26:02.95,Default,,0000,0000,0000,,All cause of X Plus One is\N0, so this one tells me that
Dialogue: 0,0:26:02.95,0:26:06.65,Default,,0000,0000,0000,,cause of X is equal to 1/2.
Dialogue: 0,0:26:07.66,0:26:13.22,Default,,0000,0000,0000,,And this one tells Maine that\Ncause of X is equal to minus
Dialogue: 0,0:26:13.22,0:26:17.93,Default,,0000,0000,0000,,one, and both of these are\Npossibilities. So I've got to
Dialogue: 0,0:26:17.93,0:26:22.64,Default,,0000,0000,0000,,solve both equations to get the\Ntotal solution to the original
Dialogue: 0,0:26:22.64,0:26:28.20,Default,,0000,0000,0000,,equation. So let's begin with\Nthis cause of X is equal to 1/2.
Dialogue: 0,0:26:28.83,0:26:35.53,Default,,0000,0000,0000,,And if you remember the range\Nof values was nought to 180
Dialogue: 0,0:26:35.53,0:26:41.66,Default,,0000,0000,0000,,degrees, so let me sketch\Ncause of X between North and
Dialogue: 0,0:26:41.66,0:26:46.69,Default,,0000,0000,0000,,180 degrees, and it looks\Nlike that zero 9180.
Dialogue: 0,0:26:47.74,0:26:53.10,Default,,0000,0000,0000,,We go across there at half and\Ncome down there and there is
Dialogue: 0,0:26:53.10,0:26:58.86,Default,,0000,0000,0000,,only one answer in the range, so\Nthat's X is equal to 60 degrees.
Dialogue: 0,0:27:00.22,0:27:06.89,Default,,0000,0000,0000,,But this one again let's sketch\Ncause of X between North and
Dialogue: 0,0:27:06.89,0:27:13.56,Default,,0000,0000,0000,,180. There and there between\Nminus one and plus one and we
Dialogue: 0,0:27:13.56,0:27:20.14,Default,,0000,0000,0000,,want cause of X equal to minus\None just at one point there and
Dialogue: 0,0:27:20.14,0:27:26.25,Default,,0000,0000,0000,,so therefore X is equal to 180\Ndegrees. So those are our two
Dialogue: 0,0:27:26.25,0:27:30.01,Default,,0000,0000,0000,,answers to the full equation\Nthat we had.
Dialogue: 0,0:27:30.06,0:27:33.82,Default,,0000,0000,0000,,So it's now have a look at
Dialogue: 0,0:27:33.82,0:27:41.22,Default,,0000,0000,0000,,three. 10 squared X is\Nequal to two sex squared X
Dialogue: 0,0:27:41.22,0:27:44.90,Default,,0000,0000,0000,,Plus One and this time will
Dialogue: 0,0:27:44.90,0:27:50.79,Default,,0000,0000,0000,,take X. To be between North and\N180 degrees. Now, the identity
Dialogue: 0,0:27:50.79,0:27:56.15,Default,,0000,0000,0000,,that we want is obviously the\None, the second one of the two
Dialogue: 0,0:27:56.15,0:28:01.09,Default,,0000,0000,0000,,that we had before. In other\Nwords, the one that tells us
Dialogue: 0,0:28:01.09,0:28:08.09,Default,,0000,0000,0000,,that sex squared X is equal to 1\N+ 10 squared X and we want to be
Dialogue: 0,0:28:08.09,0:28:14.27,Default,,0000,0000,0000,,able to take this 1 + 10 squared\Nand put it into their. So we've
Dialogue: 0,0:28:14.27,0:28:21.39,Default,,0000,0000,0000,,got 3. 10 squared\NX is equal to 2
Dialogue: 0,0:28:21.39,0:28:29.05,Default,,0000,0000,0000,,* 1 + 10 squared\NX Plus one. Multiply out
Dialogue: 0,0:28:29.05,0:28:36.71,Default,,0000,0000,0000,,this bracket. 310 squared X\Nis 2 + 210 squared
Dialogue: 0,0:28:36.71,0:28:39.01,Default,,0000,0000,0000,,X plus one.
Dialogue: 0,0:28:39.57,0:28:44.55,Default,,0000,0000,0000,,We can combine the two and the\None that will give us 3.
Dialogue: 0,0:28:45.07,0:28:51.02,Default,,0000,0000,0000,,And we can take the 210 squared\NX away from the three times
Dialogue: 0,0:28:51.02,0:28:57.44,Default,,0000,0000,0000,,squared X there. That will give\Nus 10 squared X. Now we take the
Dialogue: 0,0:28:57.44,0:29:03.85,Default,,0000,0000,0000,,square root of both sides so we\Nhave 10X is equal to plus Route
Dialogue: 0,0:29:03.85,0:29:06.14,Default,,0000,0000,0000,,3 or minus Route 3.
Dialogue: 0,0:29:07.89,0:29:11.47,Default,,0000,0000,0000,,And we need to look at each of
Dialogue: 0,0:29:11.47,0:29:15.18,Default,,0000,0000,0000,,these separately. So.
Dialogue: 0,0:29:15.77,0:29:19.04,Default,,0000,0000,0000,,Time X equals Route
Dialogue: 0,0:29:19.04,0:29:26.07,Default,,0000,0000,0000,,3. And Tan X\Nequals minus Route 3.
Dialogue: 0,0:29:26.69,0:29:34.07,Default,,0000,0000,0000,,Access to be between North and\N180, so let's have a sketch of
Dialogue: 0,0:29:34.07,0:29:40.32,Default,,0000,0000,0000,,the graph of tan between those\Nvalues, so there is 90.
Dialogue: 0,0:29:41.93,0:29:48.57,Default,,0000,0000,0000,,And there is 180 the angle whose\Ntangent is Route 3, we know.
Dialogue: 0,0:29:49.82,0:29:56.78,Default,,0000,0000,0000,,Is there at 60 so we\Nknow that X is equal to
Dialogue: 0,0:29:56.78,0:30:02.17,Default,,0000,0000,0000,,60 degrees? Here we've\Ngot minus Route 3, so
Dialogue: 0,0:30:02.17,0:30:03.73,Default,,0000,0000,0000,,again, little sketch.
Dialogue: 0,0:30:04.97,0:30:10.13,Default,,0000,0000,0000,,Between North and 180 range over\Nwhich were working here, we've
Dialogue: 0,0:30:10.13,0:30:15.76,Default,,0000,0000,0000,,got minus Route 3 go across\Nthere and down to their and
Dialogue: 0,0:30:15.76,0:30:22.32,Default,,0000,0000,0000,,symmetry says it's got to be the\Nsame as this one. Over here it's
Dialogue: 0,0:30:22.32,0:30:29.36,Default,,0000,0000,0000,,got to be the same either side.\NSo in fact if that was 60 there
Dialogue: 0,0:30:29.36,0:30:34.99,Default,,0000,0000,0000,,this must be 120 here, so X is\Nequal to 120 degrees.
Dialogue: 0,0:30:35.16,0:30:39.67,Default,,0000,0000,0000,,So far we've been working in\Ndegrees, but it makes little
Dialogue: 0,0:30:39.67,0:30:43.36,Default,,0000,0000,0000,,difference if we're actually\Nworking in radians and let's
Dialogue: 0,0:30:43.36,0:30:49.10,Default,,0000,0000,0000,,just have a look at one or two\Nexamples where in fact the range
Dialogue: 0,0:30:49.10,0:30:55.25,Default,,0000,0000,0000,,of values that we've got is in\Nradians. So if we take Tan, X is
Dialogue: 0,0:30:55.25,0:31:00.58,Default,,0000,0000,0000,,minus one and we take X to be\Nbetween plus or minus pie.
Dialogue: 0,0:31:00.87,0:31:06.16,Default,,0000,0000,0000,,Another way of looking at\Nthat would be if we were in
Dialogue: 0,0:31:06.16,0:31:10.57,Default,,0000,0000,0000,,degrees. It will be between\Nplus and minus 180. Let's
Dialogue: 0,0:31:10.57,0:31:14.10,Default,,0000,0000,0000,,sketch the graph of tangent\Nwithin that range.
Dialogue: 0,0:31:15.78,0:31:17.18,Default,,0000,0000,0000,,Up to there.
Dialogue: 0,0:31:18.06,0:31:20.73,Default,,0000,0000,0000,,That's π by 2.
Dialogue: 0,0:31:22.95,0:31:25.67,Default,,0000,0000,0000,,Up to their which is π.
Dialogue: 0,0:31:26.28,0:31:33.21,Default,,0000,0000,0000,,Minus Π\Nby 2.
Dialogue: 0,0:31:36.37,0:31:39.87,Default,,0000,0000,0000,,Their minus
Dialogue: 0,0:31:39.87,0:31:45.27,Default,,0000,0000,0000,,pie. Ton of X is\Nminus one, so somewhere
Dialogue: 0,0:31:45.27,0:31:49.19,Default,,0000,0000,0000,,across here it's going to\Nmeet the curve and we can
Dialogue: 0,0:31:49.19,0:31:50.98,Default,,0000,0000,0000,,see that means it here.
Dialogue: 0,0:31:52.12,0:31:56.06,Default,,0000,0000,0000,,And here giving us these\Nsolutions at these points. Well,
Dialogue: 0,0:31:56.06,0:32:01.58,Default,,0000,0000,0000,,we know that the angle whose\Ntangent is plus one is π by 4.
Dialogue: 0,0:32:02.33,0:32:08.96,Default,,0000,0000,0000,,So this must be pie by 4 further\Non, and so we have X is equal to
Dialogue: 0,0:32:08.96,0:32:16.37,Default,,0000,0000,0000,,pie by 2 + π by 4. That will be\N3/4 of Π or three π by 4, and
Dialogue: 0,0:32:16.37,0:32:18.32,Default,,0000,0000,0000,,this one here must be.
Dialogue: 0,0:32:19.08,0:32:25.30,Default,,0000,0000,0000,,Minus Π by 4 back there, so\Nminus π by 4.
Dialogue: 0,0:32:26.01,0:32:30.38,Default,,0000,0000,0000,,Let's take one with\Na multiple angle.
Dialogue: 0,0:32:32.24,0:32:39.10,Default,,0000,0000,0000,,So we'll have a look cause\Nof two X is equal to Route
Dialogue: 0,0:32:39.10,0:32:40.69,Default,,0000,0000,0000,,3 over 2.
Dialogue: 0,0:32:41.73,0:32:47.83,Default,,0000,0000,0000,,I will take\NX between North
Dialogue: 0,0:32:47.83,0:32:54.56,Default,,0000,0000,0000,,and 2π. Now if\NX is between North and 2π, and
Dialogue: 0,0:32:54.56,0:32:55.85,Default,,0000,0000,0000,,we've got 2X.
Dialogue: 0,0:32:56.68,0:33:01.37,Default,,0000,0000,0000,,And that means that 2X can be\Nbetween North and four π.
Dialogue: 0,0:33:02.27,0:33:07.27,Default,,0000,0000,0000,,So again, we've got to make use\Nof the periodicity.
Dialogue: 0,0:33:08.08,0:33:12.48,Default,,0000,0000,0000,,Of the graph of cosine to get a\Nsecond copy of it.
Dialogue: 0,0:33:14.38,0:33:20.70,Default,,0000,0000,0000,,So there's the first copy\Nbetween North and 2π, and now we
Dialogue: 0,0:33:20.70,0:33:27.03,Default,,0000,0000,0000,,want a second copy that goes\Nfrom 2π up till four π.
Dialogue: 0,0:33:28.38,0:33:32.88,Default,,0000,0000,0000,,We can mark these off that one\Nwill be pie by two.
Dialogue: 0,0:33:33.45,0:33:34.22,Default,,0000,0000,0000,,Pie.
Dialogue: 0,0:33:35.29,0:33:37.31,Default,,0000,0000,0000,,Three π by 2.
Dialogue: 0,0:33:38.15,0:33:46.02,Default,,0000,0000,0000,,This one will be 5 Pi by\Ntwo. This one three Pi and this
Dialogue: 0,0:33:46.02,0:33:48.83,Default,,0000,0000,0000,,one Seven π by 2.
Dialogue: 0,0:33:49.79,0:33:55.36,Default,,0000,0000,0000,,So where are we with this cost?\N2 X equals. Well, in fact we
Dialogue: 0,0:33:55.36,0:34:00.93,Default,,0000,0000,0000,,know cost to access Route 3 over\N2. We know that the angle that
Dialogue: 0,0:34:00.93,0:34:06.90,Default,,0000,0000,0000,,gives us the cosine that is\NRoute 3 over 2 is π by 6. So
Dialogue: 0,0:34:06.90,0:34:12.87,Default,,0000,0000,0000,,I'll first one is π Phi six,\Nroot 3 over 2. Up here we go
Dialogue: 0,0:34:12.87,0:34:18.45,Default,,0000,0000,0000,,across we meet the curve we come\Ndown. We know that this one here
Dialogue: 0,0:34:18.45,0:34:20.04,Default,,0000,0000,0000,,is π by 6.
Dialogue: 0,0:34:20.08,0:34:25.54,Default,,0000,0000,0000,,Let's keep going across the\Ncurves and see where we come to,
Dialogue: 0,0:34:25.54,0:34:32.36,Default,,0000,0000,0000,,what we come to one here which\Nis π by 6 short of 2π. So
Dialogue: 0,0:34:32.36,0:34:39.64,Default,,0000,0000,0000,,let me write it down as 2π -\NΠ by 6, and then again we come
Dialogue: 0,0:34:39.64,0:34:45.10,Default,,0000,0000,0000,,to one here. Symmetry suggests\Nit should be pie by 6 further
Dialogue: 0,0:34:45.10,0:34:51.02,Default,,0000,0000,0000,,on, so that's 2π + π by 6,\Nand then this one here.
Dialogue: 0,0:34:51.03,0:34:57.61,Default,,0000,0000,0000,,Is symmetry would suggest his\Npie by 6 short of four Pi,
Dialogue: 0,0:34:57.61,0:35:04.73,Default,,0000,0000,0000,,so four π - π by 6.\NSo let's do that arithmetic 2X
Dialogue: 0,0:35:04.73,0:35:06.92,Default,,0000,0000,0000,,is π by 6.
Dialogue: 0,0:35:07.53,0:35:12.95,Default,,0000,0000,0000,,Now, how many sixths are there\Nin two? Well, the answer. Is
Dialogue: 0,0:35:12.95,0:35:19.28,Default,,0000,0000,0000,,there a 12 of them and we're\Ngoing to take one of them away,
Dialogue: 0,0:35:19.28,0:35:26.06,Default,,0000,0000,0000,,so that's eleven π by 6. We're\Ngoing to now add a 6th on, so
Dialogue: 0,0:35:26.06,0:35:28.32,Default,,0000,0000,0000,,that's 13 Pi by 6.
Dialogue: 0,0:35:29.63,0:35:37.02,Default,,0000,0000,0000,,How many 6th are there in four\Nor there are 24 of them? We're
Dialogue: 0,0:35:37.02,0:35:43.89,Default,,0000,0000,0000,,going to take one away, so\Nthat's 23. Pi over 6. Now we
Dialogue: 0,0:35:43.89,0:35:51.28,Default,,0000,0000,0000,,want X, so we divide each of\Nthese by 2π by 1211 Pi by
Dialogue: 0,0:35:51.28,0:35:58.67,Default,,0000,0000,0000,,12:13, pie by 12, and 20, three\Nπ by 12, and there are our
Dialogue: 0,0:35:58.67,0:36:05.90,Default,,0000,0000,0000,,four solutions. Let's have a\Nlook at one where we've got the
Dialogue: 0,0:36:05.90,0:36:12.57,Default,,0000,0000,0000,,X divided by two rather than\Nmultiplied by two. So the sign
Dialogue: 0,0:36:12.57,0:36:18.13,Default,,0000,0000,0000,,of X over 2 is minus Route\N3 over 2.
Dialogue: 0,0:36:18.64,0:36:26.27,Default,,0000,0000,0000,,And let's take X to be\Nbetween pie and minus π. So
Dialogue: 0,0:36:26.27,0:36:33.27,Default,,0000,0000,0000,,will sketch the graph of sign\Nbetween those limited, so it's
Dialogue: 0,0:36:33.27,0:36:40.15,Default,,0000,0000,0000,,there. And their π\Nzero and minus pie.
Dialogue: 0,0:36:40.75,0:36:46.54,Default,,0000,0000,0000,,Where looking for minus three\Nover 2. Now the one thing we do
Dialogue: 0,0:36:46.54,0:36:53.21,Default,,0000,0000,0000,,know is that the angle who sign\Nis 3 over 2 is π by 3.
Dialogue: 0,0:36:53.75,0:36:59.15,Default,,0000,0000,0000,,But we want minus Route 3 over\N2, so that's down there.
Dialogue: 0,0:36:59.74,0:37:01.51,Default,,0000,0000,0000,,We go across.
Dialogue: 0,0:37:02.16,0:37:04.51,Default,,0000,0000,0000,,And we meet the curve these two
Dialogue: 0,0:37:04.51,0:37:08.72,Default,,0000,0000,0000,,points. Now this curve is\Nsymmetric with this one.
Dialogue: 0,0:37:09.23,0:37:12.07,Default,,0000,0000,0000,,So if we know that.
Dialogue: 0,0:37:12.71,0:37:14.90,Default,,0000,0000,0000,,Plus Route 3 over 2.
Dialogue: 0,0:37:15.45,0:37:21.19,Default,,0000,0000,0000,,This one was Pi by three. Then\Nwe know that this one must be
Dialogue: 0,0:37:21.19,0:37:22.83,Default,,0000,0000,0000,,minus π by 3.
Dialogue: 0,0:37:23.35,0:37:30.100,Default,,0000,0000,0000,,This one is π by three back, so\Nit's at 2π by three, so this one
Dialogue: 0,0:37:30.100,0:37:38.17,Default,,0000,0000,0000,,must be minus 2π by three, and\Nso we have X over 2 is equal
Dialogue: 0,0:37:38.17,0:37:45.34,Default,,0000,0000,0000,,to minus 2π by three and minus,\Nπ by three, but it's X that we
Dialogue: 0,0:37:45.34,0:37:52.03,Default,,0000,0000,0000,,want, so we multiply up X equals\Nminus four Pi by three and minus
Dialogue: 0,0:37:52.03,0:37:53.46,Default,,0000,0000,0000,,2π by 3.
Dialogue: 0,0:37:54.21,0:37:59.62,Default,,0000,0000,0000,,Let's just check on these\Nvalues. How do they fit with the
Dialogue: 0,0:37:59.62,0:38:05.94,Default,,0000,0000,0000,,given range? Well, this 1 - 2π\Nby three is in that given range.
Dialogue: 0,0:38:06.54,0:38:11.06,Default,,0000,0000,0000,,This one is outside, so we don't\Nwant that one.
Dialogue: 0,0:38:12.01,0:38:18.92,Default,,0000,0000,0000,,A final example here, working\Nwith the idea again of using
Dialogue: 0,0:38:18.92,0:38:24.57,Default,,0000,0000,0000,,those identities and will take 2\Ncost squared X.
Dialogue: 0,0:38:25.49,0:38:31.17,Default,,0000,0000,0000,,Plus sign X is\Nequal to 1.
Dialogue: 0,0:38:31.97,0:38:37.87,Default,,0000,0000,0000,,And we'll take X between\NNorth and 2π.
Dialogue: 0,0:38:38.78,0:38:43.06,Default,,0000,0000,0000,,We've got causes and signs,\Nso the identity that we're
Dialogue: 0,0:38:43.06,0:38:47.77,Default,,0000,0000,0000,,going to want to help us\Nwill be sine squared plus
Dialogue: 0,0:38:47.77,0:38:49.91,Default,,0000,0000,0000,,cost. Squared X equals 1.
Dialogue: 0,0:38:51.00,0:38:52.56,Default,,0000,0000,0000,,Cost squared here.
Dialogue: 0,0:38:54.50,0:38:59.72,Default,,0000,0000,0000,,Cost squared here. Let's use\Nthis identity to tell us that
Dialogue: 0,0:38:59.72,0:39:05.90,Default,,0000,0000,0000,,cost squared X is equal to 1\Nminus sign squared X and make
Dialogue: 0,0:39:05.90,0:39:08.75,Default,,0000,0000,0000,,the replacement up here for cost
Dialogue: 0,0:39:08.75,0:39:14.46,Default,,0000,0000,0000,,squared. Because that as we will\Nsee when we do it.
Dialogue: 0,0:39:14.63,0:39:22.62,Default,,0000,0000,0000,,Leads to a quadratic in sign X,\Nso it's multiply this out 2 -
Dialogue: 0,0:39:22.62,0:39:30.62,Default,,0000,0000,0000,,2 sine squared X plus sign X\Nis equal to 1 and I want
Dialogue: 0,0:39:30.62,0:39:37.47,Default,,0000,0000,0000,,it as a quadratic, so I want\Npositive square term and then
Dialogue: 0,0:39:37.47,0:39:44.89,Default,,0000,0000,0000,,the linear term and then the\Nconstant term. So I need to add.
Dialogue: 0,0:39:44.92,0:39:51.26,Default,,0000,0000,0000,,This to both sides of 0 equals 2\Nsine squared X. Adding it to
Dialogue: 0,0:39:51.26,0:39:57.60,Default,,0000,0000,0000,,both sides. Now I need to take\Nthis away minus sign X from both
Dialogue: 0,0:39:57.60,0:40:03.95,Default,,0000,0000,0000,,sides and I need to take the two\Naway from both sides. So one
Dialogue: 0,0:40:03.95,0:40:06.21,Default,,0000,0000,0000,,takeaway two is minus one.
Dialogue: 0,0:40:07.04,0:40:10.93,Default,,0000,0000,0000,,And now does this factorize?\NIt's clearly a quadratic. Let's
Dialogue: 0,0:40:10.93,0:40:16.76,Default,,0000,0000,0000,,look to see if we can make it\Nfactorize 2 sign X and sign X.
Dialogue: 0,0:40:16.76,0:40:20.66,Default,,0000,0000,0000,,Because multiplied together,\Nthese two will give Me 2 sine
Dialogue: 0,0:40:20.66,0:40:24.16,Default,,0000,0000,0000,,squared one and one because\Nmultiplied together, these two
Dialogue: 0,0:40:24.16,0:40:29.60,Default,,0000,0000,0000,,will give me one, but one of\Nthem needs to be minus. To make
Dialogue: 0,0:40:29.60,0:40:34.66,Default,,0000,0000,0000,,this a minus sign here. So I\Nthink I'll have minus there and
Dialogue: 0,0:40:34.66,0:40:39.33,Default,,0000,0000,0000,,plus there because two sign X\Ntimes by minus one gives me.
Dialogue: 0,0:40:39.39,0:40:45.60,Default,,0000,0000,0000,,Minus 2 sign X one times by sign\NX gives me sign X and if I
Dialogue: 0,0:40:45.60,0:40:50.25,Default,,0000,0000,0000,,combine sign X with minus two\Nsign XI get minus sign X.
Dialogue: 0,0:40:50.77,0:40:55.29,Default,,0000,0000,0000,,I have two numbers multiplied\Ntogether. This number 2 sign X
Dialogue: 0,0:40:55.29,0:40:59.81,Default,,0000,0000,0000,,Plus One and this number sign X\Nminus one. They multiply
Dialogue: 0,0:40:59.81,0:41:05.98,Default,,0000,0000,0000,,together to give me 0, so one or\Nboth of them must be 0. Let's
Dialogue: 0,0:41:05.98,0:41:07.21,Default,,0000,0000,0000,,write that down.
Dialogue: 0,0:41:07.94,0:41:15.60,Default,,0000,0000,0000,,2 sign X Plus One is equal to\N0 and sign X minus one is equal
Dialogue: 0,0:41:15.60,0:41:23.27,Default,,0000,0000,0000,,to 0, so this tells me that sign\Nof X is equal. To take one away
Dialogue: 0,0:41:23.27,0:41:29.97,Default,,0000,0000,0000,,from both sides and divide by\Ntwo. So sign X is minus 1/2 and
Dialogue: 0,0:41:29.97,0:41:35.24,Default,,0000,0000,0000,,this one tells me that sign X is\Nequal to 1.
Dialogue: 0,0:41:35.81,0:41:40.39,Default,,0000,0000,0000,,I'm now in a position to solve\Nthese two separate equations.
Dialogue: 0,0:41:40.91,0:41:43.36,Default,,0000,0000,0000,,So let me take this one first.
Dialogue: 0,0:41:43.98,0:41:51.12,Default,,0000,0000,0000,,Now. We were working between\NNorth and 2π, so we'll have a
Dialogue: 0,0:41:51.12,0:41:53.49,Default,,0000,0000,0000,,sketch between North and 2π.
Dialogue: 0,0:41:53.99,0:41:59.53,Default,,0000,0000,0000,,Of the sine curve and we want\Nsign X equals one. Well, there's
Dialogue: 0,0:41:59.53,0:42:05.49,Default,,0000,0000,0000,,one and there's where it meets,\Nand that's pie by two, so we can
Dialogue: 0,0:42:05.49,0:42:08.90,Default,,0000,0000,0000,,see that X is equal to pie by
Dialogue: 0,0:42:08.90,0:42:15.74,Default,,0000,0000,0000,,two. Sign X equals minus 1/2.\NAgain, the range that we've been
Dialogue: 0,0:42:15.74,0:42:21.62,Default,,0000,0000,0000,,given is between North and 2π.\NSo let's sketch between Norton
Dialogue: 0,0:42:21.62,0:42:23.22,Default,,0000,0000,0000,,2π There's 2π.
Dialogue: 0,0:42:25.45,0:42:27.09,Default,,0000,0000,0000,,Three π by 2.
Dialogue: 0,0:42:27.81,0:42:33.97,Default,,0000,0000,0000,,Pie pie by two 0 - 1/2,\Nso that's coming along between
Dialogue: 0,0:42:33.97,0:42:39.61,Default,,0000,0000,0000,,minus one and plus one that's\Ngoing to come along there.
Dialogue: 0,0:42:40.89,0:42:45.87,Default,,0000,0000,0000,,And meet the curve there and\Nthere. Now the one thing that we
Dialogue: 0,0:42:45.87,0:42:48.93,Default,,0000,0000,0000,,do know is the angle who sign is
Dialogue: 0,0:42:48.93,0:42:55.58,Default,,0000,0000,0000,,plus 1/2. Is π by 6, so we're\Nlooking at plus 1/2. It will be
Dialogue: 0,0:42:55.58,0:42:58.79,Default,,0000,0000,0000,,there and it would be pie by 6.
Dialogue: 0,0:42:59.87,0:43:06.52,Default,,0000,0000,0000,,So it's π by 6 in from there,\Nso symmetry tells us that this
Dialogue: 0,0:43:06.52,0:43:14.12,Default,,0000,0000,0000,,must be pie by 6 in from there,\Nso we've got X is equal to π
Dialogue: 0,0:43:14.12,0:43:21.72,Default,,0000,0000,0000,,+ π by 6, and symmetry tells us\Nit's pie by 6 in. From there, 2π
Dialogue: 0,0:43:21.72,0:43:23.62,Default,,0000,0000,0000,,- Π by 6.
Dialogue: 0,0:43:25.34,0:43:32.63,Default,,0000,0000,0000,,There are six sixths in pie, so\Nthat's Seven π by 6. There is
Dialogue: 0,0:43:32.63,0:43:39.41,Default,,0000,0000,0000,,1216, two Pi. We're taking one\Nof them away, so it will be
Dialogue: 0,0:43:39.41,0:43:41.49,Default,,0000,0000,0000,,11 Pi over 6.
Dialogue: 0,0:43:41.84,0:43:46.91,Default,,0000,0000,0000,,So we've shown there how to\Nsolve some trig equations.
Dialogue: 0,0:43:46.91,0:43:51.98,Default,,0000,0000,0000,,The important thing is the\Nsketch the graph. Find the
Dialogue: 0,0:43:51.98,0:43:56.54,Default,,0000,0000,0000,,initial value and then\Nworkout where the others are
Dialogue: 0,0:43:56.54,0:44:01.11,Default,,0000,0000,0000,,from the graphs. Remember,\Nthe graphs are all symmetric
Dialogue: 0,0:44:01.11,0:44:05.67,Default,,0000,0000,0000,,and they're all periodic, so\Nthey repeat themselves every
Dialogue: 0,0:44:05.67,0:44:08.20,Default,,0000,0000,0000,,2π or every 360 degrees.