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Dialogue: 0,0:00:00.96,0:00:05.62,Default,,0000,0000,0000,,In this video, we're going to be\Nlooking at basic trig identities
Dialogue: 0,0:00:05.62,0:00:09.50,Default,,0000,0000,0000,,and how to use them to solve\Ntrigonometric equations. Trig
Dialogue: 0,0:00:09.50,0:00:13.38,Default,,0000,0000,0000,,equation is an equation that\Ninvolves a trig function or
Dialogue: 0,0:00:13.38,0:00:18.42,Default,,0000,0000,0000,,functions. When we solve it,\Nwhat we do is find a value for
Dialogue: 0,0:00:18.42,0:00:22.69,Default,,0000,0000,0000,,the trig function and then find\Nthe angle that corresponds to
Dialogue: 0,0:00:22.69,0:00:26.96,Default,,0000,0000,0000,,that particular trig function.\NBut what we want to start is
Dialogue: 0,0:00:26.96,0:00:31.22,Default,,0000,0000,0000,,with the idea of a right angle\Ntriangle and go back.
Dialogue: 0,0:00:31.25,0:00:33.85,Default,,0000,0000,0000,,To the well known theorem of
Dialogue: 0,0:00:33.85,0:00:38.95,Default,,0000,0000,0000,,Pythagoras. So let's begin with\Nour right angle triangle. There
Dialogue: 0,0:00:38.95,0:00:43.96,Default,,0000,0000,0000,,the right angle in and let's\Nlabel the sides and the angles.
Dialogue: 0,0:00:43.96,0:00:45.62,Default,,0000,0000,0000,,So will have this.
Dialogue: 0,0:00:46.14,0:00:50.92,Default,,0000,0000,0000,,As a side of length a,\Nthis is a side of length
Dialogue: 0,0:00:50.92,0:00:54.90,Default,,0000,0000,0000,,B and the hypotenuse. The\Nside that is opposite the
Dialogue: 0,0:00:54.90,0:00:56.89,Default,,0000,0000,0000,,right angle will call C.
Dialogue: 0,0:00:57.93,0:01:01.66,Default,,0000,0000,0000,,And our label, this angle here\Nthe angle a.
Dialogue: 0,0:01:02.93,0:01:07.59,Default,,0000,0000,0000,,Now Pythagoras theorem tells us\Nthat if we take the square of
Dialogue: 0,0:01:07.59,0:01:10.25,Default,,0000,0000,0000,,this side. And the square of
Dialogue: 0,0:01:10.25,0:01:15.59,Default,,0000,0000,0000,,this side. Add them together.\NWe'll get the square of this
Dialogue: 0,0:01:15.59,0:01:20.68,Default,,0000,0000,0000,,side, so Pythagoras theorem\Ntells us that A squared plus B
Dialogue: 0,0:01:20.68,0:01:23.46,Default,,0000,0000,0000,,squared is equal to C squared.
Dialogue: 0,0:01:24.25,0:01:31.18,Default,,0000,0000,0000,,Now. Let's divide throughout by\NC squared, divided every term in
Dialogue: 0,0:01:31.18,0:01:33.83,Default,,0000,0000,0000,,this equation by C squared.
Dialogue: 0,0:01:34.48,0:01:40.69,Default,,0000,0000,0000,,So we have a squared over C\Nsquared plus B squared over C
Dialogue: 0,0:01:40.69,0:01:46.43,Default,,0000,0000,0000,,squared is equal to and this\Nside we would have C squared
Dialogue: 0,0:01:46.43,0:01:51.21,Default,,0000,0000,0000,,over C squared, but that of\Ncourse is just one.
Dialogue: 0,0:01:53.12,0:02:00.44,Default,,0000,0000,0000,,Now we can rewrite A squared\Nover C squared as a oversee all
Dialogue: 0,0:02:00.44,0:02:07.75,Default,,0000,0000,0000,,squared. And we can do the\Nsame here with B squared over C
Dialogue: 0,0:02:07.75,0:02:12.82,Default,,0000,0000,0000,,squared. We can rewrite that as\NB. Oversee all squared.
Dialogue: 0,0:02:12.84,0:02:14.31,Default,,0000,0000,0000,,And it still equal to 1.
Dialogue: 0,0:02:15.45,0:02:18.61,Default,,0000,0000,0000,,Let's go back to this\Ntriangle again.
Dialogue: 0,0:02:19.86,0:02:23.76,Default,,0000,0000,0000,,What does a oversee represent?
Dialogue: 0,0:02:24.46,0:02:29.40,Default,,0000,0000,0000,,Well, a over C is the side\Nopposite to the angle a.
Dialogue: 0,0:02:30.27,0:02:32.97,Default,,0000,0000,0000,,Divided by the hypotenuse.
Dialogue: 0,0:02:34.36,0:02:39.37,Default,,0000,0000,0000,,And so opposite divided by\Nhypotenuses sign. So this, a
Dialogue: 0,0:02:39.37,0:02:46.38,Default,,0000,0000,0000,,oversee is sign of a sign of\Nthe angle A and we have to
Dialogue: 0,0:02:46.38,0:02:52.79,Default,,0000,0000,0000,,square it. Now we could write\Nit like that, sign a squared,
Dialogue: 0,0:02:52.79,0:03:00.00,Default,,0000,0000,0000,,and for the moment I will plus\Nand let's have a look at B over
Dialogue: 0,0:03:00.00,0:03:07.22,Default,,0000,0000,0000,,C or B is the adjacent side to\Nthe angle A and C is the
Dialogue: 0,0:03:07.22,0:03:11.55,Default,,0000,0000,0000,,hypotenuse, so B oversee is\Nadjacent over hypotenuse and
Dialogue: 0,0:03:11.55,0:03:18.28,Default,,0000,0000,0000,,that's cosine. So we can replace\NB oversee by cause A and we need
Dialogue: 0,0:03:18.28,0:03:20.21,Default,,0000,0000,0000,,to square it still.
Dialogue: 0,0:03:20.25,0:03:21.46,Default,,0000,0000,0000,,Equal to 1.
Dialogue: 0,0:03:22.44,0:03:25.79,Default,,0000,0000,0000,,Now this notation sign.
Dialogue: 0,0:03:26.39,0:03:31.06,Default,,0000,0000,0000,,Squared, I just said sine\Nsquared, so rather than writing
Dialogue: 0,0:03:31.06,0:03:36.20,Default,,0000,0000,0000,,it to sign a squared, which\Nmight be confused with squaring
Dialogue: 0,0:03:36.20,0:03:42.74,Default,,0000,0000,0000,,the a, let's put the square on\Nthe sign and so the notation for
Dialogue: 0,0:03:42.74,0:03:49.74,Default,,0000,0000,0000,,the sign of a times by the sign\Nof a is sine squared a written
Dialogue: 0,0:03:49.74,0:03:54.88,Default,,0000,0000,0000,,like that plus, and we use\Nexactly the same technique cost
Dialogue: 0,0:03:54.88,0:03:57.99,Default,,0000,0000,0000,,squared A. Equals 1.
Dialogue: 0,0:03:58.74,0:04:06.08,Default,,0000,0000,0000,,Now that is an identity because\Nit is true for all angles
Dialogue: 0,0:04:06.08,0:04:09.76,Default,,0000,0000,0000,,a like this in a right
Dialogue: 0,0:04:09.76,0:04:12.33,Default,,0000,0000,0000,,angle triangle. However.
Dialogue: 0,0:04:12.95,0:04:18.66,Default,,0000,0000,0000,,I could have done this for the\Ndefinitions of sine and cosine
Dialogue: 0,0:04:18.66,0:04:21.52,Default,,0000,0000,0000,,that come from a unit circle.
Dialogue: 0,0:04:23.31,0:04:24.47,Default,,0000,0000,0000,,In which case?
Dialogue: 0,0:04:25.28,0:04:30.52,Default,,0000,0000,0000,,This identity would be true for\Nall angles a no matter what
Dialogue: 0,0:04:30.52,0:04:35.77,Default,,0000,0000,0000,,their size, and that's the case.\NThis is a basic trig identity
Dialogue: 0,0:04:35.77,0:04:41.01,Default,,0000,0000,0000,,that sine squared of an angle\Nplus cost squared of an angle
Dialogue: 0,0:04:41.01,0:04:44.61,Default,,0000,0000,0000,,equals 1. It's true for
Dialogue: 0,0:04:44.61,0:04:48.22,Default,,0000,0000,0000,,all angles. What developed this
Dialogue: 0,0:04:48.22,0:04:55.10,Default,,0000,0000,0000,,identity now? To give\Nus two more basic
Dialogue: 0,0:04:55.10,0:04:58.52,Default,,0000,0000,0000,,identity's. So let's
Dialogue: 0,0:04:58.52,0:05:05.59,Default,,0000,0000,0000,,begin. With sine\Nsquared A plus cost
Dialogue: 0,0:05:05.59,0:05:09.15,Default,,0000,0000,0000,,squared A equals 1.
Dialogue: 0,0:05:10.76,0:05:15.29,Default,,0000,0000,0000,,Our basic fundamental identity,\None that you really must learn
Dialogue: 0,0:05:15.29,0:05:20.73,Default,,0000,0000,0000,,and know and come to recognize\Nevery time that you see it.
Dialogue: 0,0:05:21.81,0:05:28.33,Default,,0000,0000,0000,,What I want to do is\Ndivide everything by this term
Dialogue: 0,0:05:28.33,0:05:30.70,Default,,0000,0000,0000,,here cause squared A.
Dialogue: 0,0:05:31.60,0:05:38.22,Default,,0000,0000,0000,,So sine squared a divided\Nby Cos squared A.
Dialogue: 0,0:05:38.75,0:05:46.62,Default,,0000,0000,0000,,Plus cost squared a divided\Nby Cos squared a is
Dialogue: 0,0:05:46.62,0:05:50.56,Default,,0000,0000,0000,,one over cause squared A.
Dialogue: 0,0:05:51.08,0:05:55.13,Default,,0000,0000,0000,,Now, because I've divided\Neverything by Cos squared, this
Dialogue: 0,0:05:55.13,0:05:59.63,Default,,0000,0000,0000,,is still a true equation. Still\Nin fact an identity.
Dialogue: 0,0:06:00.23,0:06:02.37,Default,,0000,0000,0000,,So sine squared over cost
Dialogue: 0,0:06:02.37,0:06:09.53,Default,,0000,0000,0000,,squared. Will sign over. Cause\Nis tan and so this is in
Dialogue: 0,0:06:09.53,0:06:11.74,Default,,0000,0000,0000,,fact TAN squared A.
Dialogue: 0,0:06:12.11,0:06:19.99,Default,,0000,0000,0000,,Plus cost squared divided by\NCos squared is just one.
Dialogue: 0,0:06:20.77,0:06:27.92,Default,,0000,0000,0000,,Now, one over cause is sick, and\Nso we can rewrite this as sex
Dialogue: 0,0:06:27.92,0:06:31.48,Default,,0000,0000,0000,,squared A. And so we have
Dialogue: 0,0:06:31.48,0:06:38.17,Default,,0000,0000,0000,,another identity. And normally\Nwe would write this as sex
Dialogue: 0,0:06:38.17,0:06:45.53,Default,,0000,0000,0000,,squared a is 1 + 10\Nsquared a. So there's our second
Dialogue: 0,0:06:45.53,0:06:49.82,Default,,0000,0000,0000,,basic fundamental identity\Nthat's derived directly from
Dialogue: 0,0:06:49.82,0:06:52.89,Default,,0000,0000,0000,,sine squared plus cost squared
Dialogue: 0,0:06:52.89,0:06:59.50,Default,,0000,0000,0000,,is one. Well, if we\Ncan divide this equation by Cos
Dialogue: 0,0:06:59.50,0:07:04.85,Default,,0000,0000,0000,,squared, surely we can do the\Nsame thing but with sine
Dialogue: 0,0:07:04.85,0:07:10.68,Default,,0000,0000,0000,,squared. So we can divide the\Nwhole of this equation by sine
Dialogue: 0,0:07:10.68,0:07:16.27,Default,,0000,0000,0000,,squared. We start again by\Nwriting down our basic
Dialogue: 0,0:07:16.27,0:07:21.35,Default,,0000,0000,0000,,fundamental identity. Sine\Nsquared plus cost squared is
Dialogue: 0,0:07:21.35,0:07:27.07,Default,,0000,0000,0000,,one. And as we said,\Ninstead of dividing everything
Dialogue: 0,0:07:27.07,0:07:32.50,Default,,0000,0000,0000,,by Cos squared, we're going to\Ndivide everything by sine
Dialogue: 0,0:07:32.50,0:07:39.32,Default,,0000,0000,0000,,squared. So that we\Nhave everything in the
Dialogue: 0,0:07:39.32,0:07:46.48,Default,,0000,0000,0000,,identity divided by sine\Nsquared. So it's still
Dialogue: 0,0:07:46.48,0:07:53.64,Default,,0000,0000,0000,,true for all angles.\NA sine squared divided
Dialogue: 0,0:07:53.64,0:07:57.22,Default,,0000,0000,0000,,by sine squared. That's
Dialogue: 0,0:07:57.22,0:07:59.48,Default,,0000,0000,0000,,just one. House.
Dialogue: 0,0:08:00.19,0:08:05.87,Default,,0000,0000,0000,,Now we've. 'cause squared\Ndivided by sine squared. So
Dialogue: 0,0:08:05.87,0:08:12.14,Default,,0000,0000,0000,,we've caused divided by sign all\Nsquared and cons divided by sign
Dialogue: 0,0:08:12.14,0:08:13.70,Default,,0000,0000,0000,,is just caught.
Dialogue: 0,0:08:14.27,0:08:16.55,Default,,0000,0000,0000,,So that is cot squared.
Dialogue: 0,0:08:17.14,0:08:22.73,Default,,0000,0000,0000,,A. Equals and here with\None over sine squared.
Dialogue: 0,0:08:23.24,0:08:29.91,Default,,0000,0000,0000,,One over sine is kosach\Nand so one over sine
Dialogue: 0,0:08:29.91,0:08:32.58,Default,,0000,0000,0000,,squared is cosec squared.
Dialogue: 0,0:08:32.59,0:08:38.42,Default,,0000,0000,0000,,And so there we have our\Nthird fundamental identity, one
Dialogue: 0,0:08:38.42,0:08:41.34,Default,,0000,0000,0000,,plus Scott squared is cosec
Dialogue: 0,0:08:41.34,0:08:48.07,Default,,0000,0000,0000,,squared. So we've\Nnow got three basic
Dialogue: 0,0:08:48.07,0:08:55.42,Default,,0000,0000,0000,,fundamental identity's. I just\Nwrite them down here in
Dialogue: 0,0:08:55.42,0:08:59.25,Default,,0000,0000,0000,,this corner sign square day.
Dialogue: 0,0:08:59.25,0:09:01.02,Default,,0000,0000,0000,,Cost Square day.
Dialogue: 0,0:09:01.56,0:09:02.51,Default,,0000,0000,0000,,Is one.
Dialogue: 0,0:09:03.12,0:09:10.90,Default,,0000,0000,0000,,1. Post\Nand squared a is sex
Dialogue: 0,0:09:10.90,0:09:18.61,Default,,0000,0000,0000,,squared A and one plus\Ncot? Squared a? Is cosec
Dialogue: 0,0:09:18.61,0:09:25.13,Default,,0000,0000,0000,,squared A? Now the use that\Nwe're going to make of these is
Dialogue: 0,0:09:25.13,0:09:28.28,Default,,0000,0000,0000,,to help us solve particular\Nkinds of trigonometric
Dialogue: 0,0:09:28.28,0:09:33.82,Default,,0000,0000,0000,,equations. So first of all,\Nlet's look at this one 2.
Dialogue: 0,0:09:35.05,0:09:37.26,Default,,0000,0000,0000,,10 squared X.
Dialogue: 0,0:09:37.99,0:09:42.62,Default,,0000,0000,0000,,Is equal to sex\Nsquared X.
Dialogue: 0,0:09:44.28,0:09:49.03,Default,,0000,0000,0000,,What we need to do is looking\Nat this equation related to
Dialogue: 0,0:09:49.03,0:09:52.99,Default,,0000,0000,0000,,one of these three identity's\Nand it's fairly obvious that
Dialogue: 0,0:09:52.99,0:09:54.58,Default,,0000,0000,0000,,this is the one.
Dialogue: 0,0:09:55.86,0:10:01.14,Default,,0000,0000,0000,,So what we need to do then is\Nget everything in terms of
Dialogue: 0,0:10:01.14,0:10:03.64,Default,,0000,0000,0000,,either tans. Or sex.
Dialogue: 0,0:10:04.69,0:10:11.01,Default,,0000,0000,0000,,Well, our identity says that's X\Nsquared is equal to 1 + 10
Dialogue: 0,0:10:11.01,0:10:16.84,Default,,0000,0000,0000,,squared. So let's replace the\Nsex squared here by 1 + 10
Dialogue: 0,0:10:16.84,0:10:23.16,Default,,0000,0000,0000,,squared. So we have two 10\Nsquared X is equal to 1 plus.
Dialogue: 0,0:10:23.68,0:10:26.33,Default,,0000,0000,0000,,10 squared X.
Dialogue: 0,0:10:27.29,0:10:33.04,Default,,0000,0000,0000,,And now we can take 10 squared\Naway from each side, which will
Dialogue: 0,0:10:33.04,0:10:38.34,Default,,0000,0000,0000,,leave us with one 10 squared\Nthis side and equals 1 there.
Dialogue: 0,0:10:39.47,0:10:43.88,Default,,0000,0000,0000,,So now we can take the square\Nroot of both sides, so Tan X is
Dialogue: 0,0:10:43.88,0:10:47.70,Default,,0000,0000,0000,,equal to 1. And let's not forget\Nwhen we take a square root.
Dialogue: 0,0:10:48.24,0:10:53.98,Default,,0000,0000,0000,,We've got 2 answers, plus or\Nminus one in this case.
Dialogue: 0,0:10:54.89,0:10:58.42,Default,,0000,0000,0000,,Now, the one thing that we\Ndidn't specify at the beginning
Dialogue: 0,0:10:58.42,0:11:02.59,Default,,0000,0000,0000,,of this question was what was\Nthe range of values that we were
Dialogue: 0,0:11:02.59,0:11:04.52,Default,,0000,0000,0000,,going to be working with 4X.
Dialogue: 0,0:11:05.24,0:11:08.68,Default,,0000,0000,0000,,Well, since we didn't specify at\Nthe beginning, I think we're
Dialogue: 0,0:11:08.68,0:11:13.06,Default,,0000,0000,0000,,entitled to put in any range of\Nvalues that we want. So for the
Dialogue: 0,0:11:13.06,0:11:16.82,Default,,0000,0000,0000,,moment, let's say that we're\Ngoing to look at this for X.
Dialogue: 0,0:11:17.36,0:11:23.74,Default,,0000,0000,0000,,Between equal to 0 but\Nless than 2π.
Dialogue: 0,0:11:24.83,0:11:30.68,Default,,0000,0000,0000,,Sotan axes one or minus one.\NLet's sketch the graph.
Dialogue: 0,0:11:31.76,0:11:36.76,Default,,0000,0000,0000,,Of tanks between North and 2π,\Nso it looks like that.
Dialogue: 0,0:11:37.45,0:11:43.44,Default,,0000,0000,0000,,Asymptotes. Like that asymptotes\Nand like that.
Dialogue: 0,0:11:44.06,0:11:49.71,Default,,0000,0000,0000,,This is 2π this one we know\Nis π by 2.
Dialogue: 0,0:11:50.34,0:11:56.76,Default,,0000,0000,0000,,This where it crosses the X axis\Nis π, and this one here is 3 Pi
Dialogue: 0,0:11:56.76,0:12:00.68,Default,,0000,0000,0000,,by two. 10 X equals
Dialogue: 0,0:12:00.68,0:12:04.77,Default,,0000,0000,0000,,1. That's one of those\Nspecial angles.
Dialogue: 0,0:12:05.85,0:12:11.66,Default,,0000,0000,0000,,45 degrees if we were working in\Ndegrees or pie by 4 radians. If
Dialogue: 0,0:12:11.66,0:12:17.47,Default,,0000,0000,0000,,we're working in radians. So for\Nthe one bit we want Thai by 4,
Dialogue: 0,0:12:17.47,0:12:23.70,Default,,0000,0000,0000,,but where else do we want to be?\NHere's one, and if we go across
Dialogue: 0,0:12:23.70,0:12:27.02,Default,,0000,0000,0000,,the tan graph we can see we meet
Dialogue: 0,0:12:27.02,0:12:33.25,Default,,0000,0000,0000,,it here. That's the pie by 4\Nand we meet it here.
Dialogue: 0,0:12:34.43,0:12:40.37,Default,,0000,0000,0000,,And that is going to be in there\Nhalfway between pie and three Pi
Dialogue: 0,0:12:40.37,0:12:47.15,Default,,0000,0000,0000,,by two, and so that is going to\Nbe at five π by 4. So there's
Dialogue: 0,0:12:47.15,0:12:48.42,Default,,0000,0000,0000,,our second answer.
Dialogue: 0,0:12:49.03,0:12:54.50,Default,,0000,0000,0000,,Coming from one and then we've\Ngot the minus one, so let's go
Dialogue: 0,0:12:54.50,0:12:56.19,Default,,0000,0000,0000,,across at minus one.
Dialogue: 0,0:12:57.45,0:12:58.75,Default,,0000,0000,0000,,Till we meet the graph.
Dialogue: 0,0:13:00.04,0:13:02.04,Default,,0000,0000,0000,,There and there.
Dialogue: 0,0:13:02.66,0:13:09.16,Default,,0000,0000,0000,,This is half way between pie by\Ntwo and Π, and so that's going
Dialogue: 0,0:13:09.16,0:13:11.94,Default,,0000,0000,0000,,to be three π by 4.
Dialogue: 0,0:13:12.72,0:13:18.67,Default,,0000,0000,0000,,And this one is halfway between\Nthree Pi by two an 2π and so
Dialogue: 0,0:13:18.67,0:13:23.77,Default,,0000,0000,0000,,that's going to be 7 Pi by 4,\Nand so there are.
Dialogue: 0,0:13:25.55,0:13:28.27,Default,,0000,0000,0000,,Four solutions to this question.
Dialogue: 0,0:13:29.61,0:13:35.08,Default,,0000,0000,0000,,Let's take another example and\Nthis time I'm going to take one
Dialogue: 0,0:13:35.08,0:13:40.10,Default,,0000,0000,0000,,that will make use of one more\Nof these particular basic
Dialogue: 0,0:13:40.10,0:13:46.73,Default,,0000,0000,0000,,identity's. So two\Nsine squared
Dialogue: 0,0:13:46.73,0:13:50.02,Default,,0000,0000,0000,,X. Close
Dialogue: 0,0:13:50.02,0:13:55.14,Default,,0000,0000,0000,,call sex.\NEquals 1.
Dialogue: 0,0:13:56.23,0:13:58.60,Default,,0000,0000,0000,,Now this has got sine squared's
Dialogue: 0,0:13:58.60,0:14:01.37,Default,,0000,0000,0000,,in it. And the cause?
Dialogue: 0,0:14:01.93,0:14:05.23,Default,,0000,0000,0000,,Well, fairly obviously, I think\Nwe ought to be using sine
Dialogue: 0,0:14:05.23,0:14:06.73,Default,,0000,0000,0000,,squared plus cost. Squared is
Dialogue: 0,0:14:06.73,0:14:13.28,Default,,0000,0000,0000,,one. But what do we replace? Do\Nwe try and replace the cause or
Dialogue: 0,0:14:13.28,0:14:18.95,Default,,0000,0000,0000,,do we try and replace the sign?\NWe've got a choice. Well, the
Dialogue: 0,0:14:18.95,0:14:22.87,Default,,0000,0000,0000,,identity says sign squared plus\Ncost squared equals 1.
Dialogue: 0,0:14:23.60,0:14:29.33,Default,,0000,0000,0000,,So if it's sine squared that's\Nin the identity, then perhaps
Dialogue: 0,0:14:29.33,0:14:35.06,Default,,0000,0000,0000,,it's the sine squared that we\Nought to replace. So let's
Dialogue: 0,0:14:35.06,0:14:38.71,Default,,0000,0000,0000,,make that replacement instead\Nof sine squared.
Dialogue: 0,0:14:39.74,0:14:46.07,Default,,0000,0000,0000,,Be'cause sine squared plus cost\Nsquared is one sign. Squared
Dialogue: 0,0:14:46.07,0:14:53.67,Default,,0000,0000,0000,,must be 1 minus Cos squared.\NSo in there will write 1
Dialogue: 0,0:14:53.67,0:14:59.36,Default,,0000,0000,0000,,minus Cos squared X plus cause\NX equals 1.
Dialogue: 0,0:15:00.62,0:15:07.40,Default,,0000,0000,0000,,Multiply out the brackets\N2 - 2 cost
Dialogue: 0,0:15:07.40,0:15:13.32,Default,,0000,0000,0000,,squared X plus cause\NX equals 1.
Dialogue: 0,0:15:14.46,0:15:19.96,Default,,0000,0000,0000,,Well, if we simplify this, what\Nwe're going to end up with is a
Dialogue: 0,0:15:19.96,0:15:24.28,Default,,0000,0000,0000,,quadratic equation where the\Nvariable is going to be cause X.
Dialogue: 0,0:15:24.28,0:15:29.39,Default,,0000,0000,0000,,So let's move this term minus\Ntwo cost squared X over to this
Dialogue: 0,0:15:29.39,0:15:34.50,Default,,0000,0000,0000,,side of the equation. By adding\Ntwo cost squared X to each side
Dialogue: 0,0:15:34.50,0:15:40.40,Default,,0000,0000,0000,,so that we get it positive at\Nthis side. And So what I want to
Dialogue: 0,0:15:40.40,0:15:44.72,Default,,0000,0000,0000,,end up with is an equation\Nthat's equal to 0, so.
Dialogue: 0,0:15:44.78,0:15:51.81,Default,,0000,0000,0000,,Equals 0 at this to both\Nsides. To cost squared X, take
Dialogue: 0,0:15:51.81,0:15:58.26,Default,,0000,0000,0000,,this away from both sides,\Nbecause that's plus cause X, so
Dialogue: 0,0:15:58.26,0:16:00.02,Default,,0000,0000,0000,,minus cause X.
Dialogue: 0,0:16:00.65,0:16:06.65,Default,,0000,0000,0000,,Take the two away from both\Nsides, so that's one. Take away
Dialogue: 0,0:16:06.65,0:16:09.15,Default,,0000,0000,0000,,these two is minus one.
Dialogue: 0,0:16:10.03,0:16:15.79,Default,,0000,0000,0000,,As we said before, this is\Nnow just a quadratic.
Dialogue: 0,0:16:16.30,0:16:20.23,Default,,0000,0000,0000,,So let's see if we can
Dialogue: 0,0:16:20.23,0:16:26.73,Default,,0000,0000,0000,,factorize it. Remembering that\Nthe variable is cause X, well we
Dialogue: 0,0:16:26.73,0:16:32.46,Default,,0000,0000,0000,,want two numbers that will\Nmultiply together is to give us
Dialogue: 0,0:16:32.46,0:16:39.23,Default,,0000,0000,0000,,2 cost squared X, which suggests\Nperhaps two Cos X and cause X.
Dialogue: 0,0:16:40.17,0:16:44.48,Default,,0000,0000,0000,,We want two numbers that will\Nmultiply together to give us
Dialogue: 0,0:16:44.48,0:16:49.19,Default,,0000,0000,0000,,minus one. Well, let's put ones\Nin for the moment and worry
Dialogue: 0,0:16:49.19,0:16:50.75,Default,,0000,0000,0000,,about the sign now.
Dialogue: 0,0:16:51.49,0:16:57.56,Default,,0000,0000,0000,,If I take 2 cause X times\Nby one here, I will get
Dialogue: 0,0:16:57.56,0:16:59.43,Default,,0000,0000,0000,,just to cause X.
Dialogue: 0,0:17:00.83,0:17:07.09,Default,,0000,0000,0000,,If I take 1 by cause X here I\Nwill get just cause X and I want
Dialogue: 0,0:17:07.09,0:17:11.87,Default,,0000,0000,0000,,to end up with minus Cos X,\Nwhich means I've really got to
Dialogue: 0,0:17:11.87,0:17:14.08,Default,,0000,0000,0000,,take away the result of doing
Dialogue: 0,0:17:14.08,0:17:20.33,Default,,0000,0000,0000,,this multiplication. So the\Nminus sign there and a plus side
Dialogue: 0,0:17:20.33,0:17:28.18,Default,,0000,0000,0000,,there. So what does this\Ntell us? If this expression is
Dialogue: 0,0:17:28.18,0:17:35.91,Default,,0000,0000,0000,,equal to 0, then either 2\Ncalls X plus one equals 0
Dialogue: 0,0:17:35.91,0:17:37.84,Default,,0000,0000,0000,,or cause X.
Dialogue: 0,0:17:38.45,0:17:44.13,Default,,0000,0000,0000,,Minus one equals 0. This gives\Nus a nice little equation that
Dialogue: 0,0:17:44.13,0:17:50.28,Default,,0000,0000,0000,,says cause X is equal to. I'll\Ntake one away from both sides,
Dialogue: 0,0:17:50.28,0:17:56.90,Default,,0000,0000,0000,,so that's minus one and divide\Nboth sides by two, so cause X is
Dialogue: 0,0:17:56.90,0:17:59.26,Default,,0000,0000,0000,,equal to minus 1/2 or.
Dialogue: 0,0:17:59.89,0:18:03.65,Default,,0000,0000,0000,,Cause X is equal to 1.
Dialogue: 0,0:18:04.35,0:18:09.80,Default,,0000,0000,0000,,And so in order to solve this\Nequation at the top, I've now
Dialogue: 0,0:18:09.80,0:18:14.41,Default,,0000,0000,0000,,reduced it to solving these two\Nmuch simpler equations at the
Dialogue: 0,0:18:14.41,0:18:19.70,Default,,0000,0000,0000,,bottom. So I'll turn over the\Npage now and take these two with
Dialogue: 0,0:18:19.70,0:18:27.14,Default,,0000,0000,0000,,me. So cause X\Nis equal to minus
Dialogue: 0,0:18:27.14,0:18:34.30,Default,,0000,0000,0000,,1/2. All.\NKohl's X is equal to wall.
Dialogue: 0,0:18:35.27,0:18:40.57,Default,,0000,0000,0000,,Now again, when we start at\Nsolving this, we did not have a
Dialogue: 0,0:18:40.57,0:18:45.88,Default,,0000,0000,0000,,range of values for Cos X. So\Nlet's say that again we're going
Dialogue: 0,0:18:45.88,0:18:50.77,Default,,0000,0000,0000,,to work with this range of\Nvalues. X is greater than or
Dialogue: 0,0:18:50.77,0:18:56.89,Default,,0000,0000,0000,,equal to 0, but less than two\NPi. What we need to do first is
Dialogue: 0,0:18:56.89,0:19:02.61,Default,,0000,0000,0000,,sketch the graph of Cos X in\Nthat range. So the graph of Cos
Dialogue: 0,0:19:02.61,0:19:05.46,Default,,0000,0000,0000,,X in that range looks like that.
Dialogue: 0,0:19:07.38,0:19:09.93,Default,,0000,0000,0000,,This is π by 2.
Dialogue: 0,0:19:10.79,0:19:13.14,Default,,0000,0000,0000,,This is pie.
Dialogue: 0,0:19:14.16,0:19:17.67,Default,,0000,0000,0000,,Three Pi by two
Dialogue: 0,0:19:17.67,0:19:25.50,Default,,0000,0000,0000,,and 2π. This is one\Non the Y axis and minus one on
Dialogue: 0,0:19:25.50,0:19:31.33,Default,,0000,0000,0000,,the X axis. So what are our\Nvalues? Well, let's take this
Dialogue: 0,0:19:31.33,0:19:36.67,Default,,0000,0000,0000,,equation first cause X equals\None. We go across at one.
Dialogue: 0,0:19:37.61,0:19:39.36,Default,,0000,0000,0000,,We've got this value here.
Dialogue: 0,0:19:40.00,0:19:42.54,Default,,0000,0000,0000,,X equals 0.
Dialogue: 0,0:19:43.42,0:19:46.80,Default,,0000,0000,0000,,And this value here X equals 2π,
Dialogue: 0,0:19:46.80,0:19:52.57,Default,,0000,0000,0000,,but. This here says X is\Nstrictly less than 2π, so
Dialogue: 0,0:19:52.57,0:19:54.38,Default,,0000,0000,0000,,if I included it.
Dialogue: 0,0:19:55.50,0:20:00.36,Default,,0000,0000,0000,,Be right across it out because\Nit's not within the range of
Dialogue: 0,0:20:00.36,0:20:05.22,Default,,0000,0000,0000,,values. Let's now have a look\Nat this cause X equals minus
Dialogue: 0,0:20:05.22,0:20:10.08,Default,,0000,0000,0000,,1/2. Well minus 1/2 is there.\NSo let's go across and see
Dialogue: 0,0:20:10.08,0:20:13.32,Default,,0000,0000,0000,,where this meets the graph\Nthere and there.
Dialogue: 0,0:20:14.43,0:20:20.60,Default,,0000,0000,0000,,Right, this again, is connected\Nwith one of those very special
Dialogue: 0,0:20:20.60,0:20:26.72,Default,,0000,0000,0000,,angles. If concept X had\Nbeen equal to 1/2, then
Dialogue: 0,0:20:26.72,0:20:32.41,Default,,0000,0000,0000,,X would be equal to 60\Ndegrees or pie by three.
Dialogue: 0,0:20:33.94,0:20:37.03,Default,,0000,0000,0000,,That's about there.
Dialogue: 0,0:20:37.90,0:20:40.81,Default,,0000,0000,0000,,These curves are symmetric, so
Dialogue: 0,0:20:40.81,0:20:48.33,Default,,0000,0000,0000,,this one. Instead of being\Npie by three from there is π
Dialogue: 0,0:20:48.33,0:20:51.14,Default,,0000,0000,0000,,by three back from pie.
Dialogue: 0,0:20:51.85,0:20:58.08,Default,,0000,0000,0000,,And so this tells us that X is\Nequal to 2π by 3.
Dialogue: 0,0:20:58.68,0:21:05.34,Default,,0000,0000,0000,,Or it's pie by three on from\Nthere, which gives us four Pi by
Dialogue: 0,0:21:05.34,0:21:11.06,Default,,0000,0000,0000,,three. So there we've got our\Nanswers. Two of them there, and
Dialogue: 0,0:21:11.06,0:21:12.96,Default,,0000,0000,0000,,one of them there.
Dialogue: 0,0:21:14.19,0:21:23.15,Default,,0000,0000,0000,,So.\NLet's take another example.
Dialogue: 0,0:21:23.15,0:21:29.82,Default,,0000,0000,0000,,Three cop squared X\Nis equal to Cosec
Dialogue: 0,0:21:29.82,0:21:33.41,Default,,0000,0000,0000,,X. Minus one.
Dialogue: 0,0:21:34.11,0:21:41.13,Default,,0000,0000,0000,,So the identity that we want\Nis the one that talks to
Dialogue: 0,0:21:41.13,0:21:47.56,Default,,0000,0000,0000,,us about cot squared and Cosec\Nsquared. But which term should
Dialogue: 0,0:21:47.56,0:21:54.00,Default,,0000,0000,0000,,we replace now? Let's recall the\Nidentity is one plus cot
Dialogue: 0,0:21:54.00,0:21:57.51,Default,,0000,0000,0000,,squared. X is cosec squared X.
Dialogue: 0,0:21:58.13,0:22:02.61,Default,,0000,0000,0000,,Well, as we saw in the last\Nexample, we want to arrive at a
Dialogue: 0,0:22:02.61,0:22:06.13,Default,,0000,0000,0000,,quadratic that we can factorise\Nit therefore makes no sense to
Dialogue: 0,0:22:06.13,0:22:10.61,Default,,0000,0000,0000,,try and substitute for the Cosec\N'cause to do that we have to get
Dialogue: 0,0:22:10.61,0:22:11.89,Default,,0000,0000,0000,,square roots in it.
Dialogue: 0,0:22:12.66,0:22:17.92,Default,,0000,0000,0000,,But if we substitute for the cot\Nsquared, we can do so much
Dialogue: 0,0:22:17.92,0:22:23.80,Default,,0000,0000,0000,,better. Because we will just\Nhave a direct substitution that
Dialogue: 0,0:22:23.80,0:22:29.06,Default,,0000,0000,0000,,will involve cosec squared and\Nhopefully get a quadratic. So
Dialogue: 0,0:22:29.06,0:22:34.32,Default,,0000,0000,0000,,instead of caught squared will\Nreplace it. By changing this
Dialogue: 0,0:22:34.32,0:22:40.11,Default,,0000,0000,0000,,around, that tells us that\Ncaught squared X is equal to
Dialogue: 0,0:22:40.11,0:22:45.90,Default,,0000,0000,0000,,cosine X squared X minus one, so\Nthat will be 3.
Dialogue: 0,0:22:46.67,0:22:53.61,Default,,0000,0000,0000,,Cosec squared X minus\None is equal to
Dialogue: 0,0:22:53.61,0:22:57.07,Default,,0000,0000,0000,,cosec X minus one.
Dialogue: 0,0:22:58.05,0:23:00.12,Default,,0000,0000,0000,,Multiply out the brackets.
Dialogue: 0,0:23:00.77,0:23:07.56,Default,,0000,0000,0000,,3. Cosec squared\NX minus three. Don't forget when
Dialogue: 0,0:23:07.56,0:23:12.00,Default,,0000,0000,0000,,you multiply out brackets, you\Nmust multiply everything inside
Dialogue: 0,0:23:12.00,0:23:17.92,Default,,0000,0000,0000,,the bracket by what's outside,\Nso we've got to have the three
Dialogue: 0,0:23:17.92,0:23:23.34,Default,,0000,0000,0000,,times by the minus one there\Nequals cosec X minus one.
Dialogue: 0,0:23:24.26,0:23:28.11,Default,,0000,0000,0000,,Let's get everything on one side\Nof the equation so it says
Dialogue: 0,0:23:28.11,0:23:33.44,Default,,0000,0000,0000,,equals 0. Keep the square term\Nto be the positive term.
Dialogue: 0,0:23:33.97,0:23:37.97,Default,,0000,0000,0000,,That makes factorization\Neasier, so free KOs X
Dialogue: 0,0:23:37.97,0:23:42.47,Default,,0000,0000,0000,,squared X takeaway. This\Ncode set from each side.
Dialogue: 0,0:23:43.60,0:23:49.22,Default,,0000,0000,0000,,And now I've minus three here,\Nand I've minus one here. I want
Dialogue: 0,0:23:49.22,0:23:56.13,Default,,0000,0000,0000,,to take the minus one over to\Nthat site, so I have to add 1 to
Dialogue: 0,0:23:56.13,0:24:00.45,Default,,0000,0000,0000,,each side, so minus three plus\NOne is minus 2.
Dialogue: 0,0:24:00.47,0:24:01.58,Default,,0000,0000,0000,,Equals 0.
Dialogue: 0,0:24:02.77,0:24:06.14,Default,,0000,0000,0000,,Now. Factorise
Dialogue: 0,0:24:06.14,0:24:12.45,Default,,0000,0000,0000,,this equation. The\Nvariable is kosach, so we're
Dialogue: 0,0:24:12.45,0:24:15.75,Default,,0000,0000,0000,,going to have three cosec in
Dialogue: 0,0:24:15.75,0:24:21.52,Default,,0000,0000,0000,,here. And cosec X in there\Nbecause that kinds by that will
Dialogue: 0,0:24:21.52,0:24:26.50,Default,,0000,0000,0000,,give us that term there three\Ncosec squared X. And now I've
Dialogue: 0,0:24:26.50,0:24:29.41,Default,,0000,0000,0000,,got the minus two to deal with.
Dialogue: 0,0:24:30.00,0:24:35.24,Default,,0000,0000,0000,,Well to itself is 2 times by\None. If I put the two in here
Dialogue: 0,0:24:35.24,0:24:39.42,Default,,0000,0000,0000,,then I'm going to multiply the\Ntwo by the three, and that's
Dialogue: 0,0:24:39.42,0:24:43.96,Default,,0000,0000,0000,,going to give me 6, which is a\Nvery big number in Association
Dialogue: 0,0:24:43.96,0:24:48.15,Default,,0000,0000,0000,,with the cosack. I only want\Nminus one cosec, so let's put
Dialogue: 0,0:24:48.15,0:24:53.03,Default,,0000,0000,0000,,the two in there on the one in\Nthere. Now I've got to balance
Dialogue: 0,0:24:53.03,0:24:55.13,Default,,0000,0000,0000,,the signs I want, minus two
Dialogue: 0,0:24:55.13,0:25:00.64,Default,,0000,0000,0000,,here. So one of these has got to\Nbe negative and I see that
Dialogue: 0,0:25:00.64,0:25:04.40,Default,,0000,0000,0000,,what's going to make the\Ndecision for me. Is this minus
Dialogue: 0,0:25:04.40,0:25:09.44,Default,,0000,0000,0000,,cosec X? So I need the bigger\Nbit in size to be negative.
Dialogue: 0,0:25:10.03,0:25:16.16,Default,,0000,0000,0000,,Which seems to me that the three\Ncosec X has got to go with us
Dialogue: 0,0:25:16.16,0:25:21.07,Default,,0000,0000,0000,,minus one, so three cosec times\Nby minus one is minus three
Dialogue: 0,0:25:21.07,0:25:25.67,Default,,0000,0000,0000,,kosek. And then I've got +2\Nkocek there gives me the minus
Dialogue: 0,0:25:25.67,0:25:27.94,Default,,0000,0000,0000,,the single cosec that I want.
Dialogue: 0,0:25:28.53,0:25:34.34,Default,,0000,0000,0000,,Solving a quadratic two brackets\Nmultiplied together equal 0.
Dialogue: 0,0:25:34.34,0:25:40.80,Default,,0000,0000,0000,,That means one of these brackets\Nor the other one.
Dialogue: 0,0:25:41.51,0:25:44.95,Default,,0000,0000,0000,,Has got to be
Dialogue: 0,0:25:44.95,0:25:48.27,Default,,0000,0000,0000,,equal to. 0.
Dialogue: 0,0:25:50.24,0:25:57.34,Default,,0000,0000,0000,,So. But this one I\Ncan take two away from each side
Dialogue: 0,0:25:57.34,0:26:04.14,Default,,0000,0000,0000,,and divide by three. So we have\Ncosec X is equal to minus two
Dialogue: 0,0:26:04.14,0:26:06.56,Default,,0000,0000,0000,,over three or minus 2/3.
Dialogue: 0,0:26:07.07,0:26:09.75,Default,,0000,0000,0000,,And here kosek.
Dialogue: 0,0:26:10.30,0:26:16.07,Default,,0000,0000,0000,,X is equal to 1, so\Nagain we've reduced.
Dialogue: 0,0:26:16.80,0:26:23.55,Default,,0000,0000,0000,,An equation like this to\Nsolving two much smaller, much
Dialogue: 0,0:26:23.55,0:26:29.94,Default,,0000,0000,0000,,simpler equations. So taking\Nthese two over the
Dialogue: 0,0:26:29.94,0:26:35.80,Default,,0000,0000,0000,,page cosec X is\Nminus 2/3 or?
Dialogue: 0,0:26:36.50,0:26:40.22,Default,,0000,0000,0000,,Cosec X is one.
Dialogue: 0,0:26:41.32,0:26:48.42,Default,,0000,0000,0000,,Now what is cosec? Cosack\Nis one over sine X.
Dialogue: 0,0:26:48.47,0:26:52.26,Default,,0000,0000,0000,,So let's write
Dialogue: 0,0:26:52.26,0:26:58.76,Default,,0000,0000,0000,,that down. If you just\Nlook at this one over sine X
Dialogue: 0,0:26:58.76,0:27:03.82,Default,,0000,0000,0000,,equals 1. While that can only\Nmean cynex itself is equal to 1,
Dialogue: 0,0:27:03.82,0:27:09.26,Default,,0000,0000,0000,,and if we can turn this upside\Ndown, we can do the same this
Dialogue: 0,0:27:09.26,0:27:13.93,Default,,0000,0000,0000,,side. So turning that back\Nupside down, sign X is equal to
Dialogue: 0,0:27:13.93,0:27:15.49,Default,,0000,0000,0000,,minus three over 2.
Dialogue: 0,0:27:16.65,0:27:21.78,Default,,0000,0000,0000,,Now when we began this equation\Nand we began to solve it, we
Dialogue: 0,0:27:21.78,0:27:26.92,Default,,0000,0000,0000,,didn't state a range of values\Nof X, so let's use the range
Dialogue: 0,0:27:26.92,0:27:32.06,Default,,0000,0000,0000,,again that we've been using and\Nthat is X greater than or equal
Dialogue: 0,0:27:32.06,0:27:34.42,Default,,0000,0000,0000,,to 0, but less than 2π.
Dialogue: 0,0:27:35.52,0:27:41.56,Default,,0000,0000,0000,,Let's just sketch the graph of\Ncynex over that range of values.
Dialogue: 0,0:27:42.67,0:27:46.91,Default,,0000,0000,0000,,Graph of Cynex will look like\Nthat going from North.
Dialogue: 0,0:27:47.80,0:27:49.79,Default,,0000,0000,0000,,Through pie by two.
Dialogue: 0,0:27:50.96,0:27:52.13,Default,,0000,0000,0000,,Pie.
Dialogue: 0,0:27:53.57,0:27:55.34,Default,,0000,0000,0000,,Three π by 2.
Dialogue: 0,0:27:55.94,0:28:02.14,Default,,0000,0000,0000,,And 2π and it will range between\Nplus one and minus one. So
Dialogue: 0,0:28:02.14,0:28:07.39,Default,,0000,0000,0000,,again, if we look at this\Nsolution here, we can see
Dialogue: 0,0:28:07.39,0:28:13.11,Default,,0000,0000,0000,,straightaway. We've got the one\Nsolution here at X equals π by
Dialogue: 0,0:28:13.11,0:28:19.57,Default,,0000,0000,0000,,2. Let's have a look at this\None. Sign X equals minus three
Dialogue: 0,0:28:19.57,0:28:21.67,Default,,0000,0000,0000,,over two. Well, that's here.
Dialogue: 0,0:28:22.38,0:28:24.76,Default,,0000,0000,0000,,And of course there's a problem.
Dialogue: 0,0:28:26.18,0:28:30.72,Default,,0000,0000,0000,,This doesn't mean the graph\Nanywhere. There are no values of
Dialogue: 0,0:28:30.72,0:28:33.20,Default,,0000,0000,0000,,X that will produce minus three
Dialogue: 0,0:28:33.20,0:28:38.08,Default,,0000,0000,0000,,over 2. Be'cause sign is\Ncontained between plus one
Dialogue: 0,0:28:38.08,0:28:43.08,Default,,0000,0000,0000,,and minus one. That doesn't\Nmean that we've done it
Dialogue: 0,0:28:43.08,0:28:49.08,Default,,0000,0000,0000,,wrong. All it means is that\Nthere are no values of X
Dialogue: 0,0:28:49.08,0:28:53.58,Default,,0000,0000,0000,,that come from this\Nequation, and so the only
Dialogue: 0,0:28:53.58,0:28:57.58,Default,,0000,0000,0000,,solutions are those that\Ncome from this equation
Dialogue: 0,0:28:57.58,0:28:59.08,Default,,0000,0000,0000,,here, so this.
Dialogue: 0,0:29:00.21,0:29:03.55,Default,,0000,0000,0000,,Is our answer and\Nour only answer.
Dialogue: 0,0:29:04.62,0:29:06.66,Default,,0000,0000,0000,,Will take one more.
Dialogue: 0,0:29:07.41,0:29:14.63,Default,,0000,0000,0000,,Example, this one is cost\Nsquared X minus sign squared X
Dialogue: 0,0:29:14.63,0:29:17.25,Default,,0000,0000,0000,,is equal to 0.
Dialogue: 0,0:29:18.97,0:29:22.46,Default,,0000,0000,0000,,Now we've got an identity\Nthat says cost squared plus
Dialogue: 0,0:29:22.46,0:29:25.95,Default,,0000,0000,0000,,sign squared is one, so I\Ncould choose to replace
Dialogue: 0,0:29:25.95,0:29:28.74,Default,,0000,0000,0000,,either the cost squared all\Nthe sine squared.
Dialogue: 0,0:29:30.51,0:29:35.19,Default,,0000,0000,0000,,But The reason I've chosen\Nthis example is that you can
Dialogue: 0,0:29:35.19,0:29:36.56,Default,,0000,0000,0000,,do it another way.
Dialogue: 0,0:29:37.82,0:29:40.63,Default,,0000,0000,0000,,So let's have a look at the
Dialogue: 0,0:29:40.63,0:29:47.31,Default,,0000,0000,0000,,other way. This is cost\Nsquared minus sign squared. So
Dialogue: 0,0:29:47.31,0:29:53.86,Default,,0000,0000,0000,,in algebra terms it's the\Ndifference of two squares. It's
Dialogue: 0,0:29:53.86,0:30:00.41,Default,,0000,0000,0000,,A squared minus B squared and\Nthat has a standard
Dialogue: 0,0:30:00.41,0:30:03.03,Default,,0000,0000,0000,,factorization of A-B A+B.
Dialogue: 0,0:30:03.77,0:30:09.84,Default,,0000,0000,0000,,So this factorizes\Nas cause X
Dialogue: 0,0:30:09.84,0:30:12.87,Default,,0000,0000,0000,,minus sign X.
Dialogue: 0,0:30:12.87,0:30:19.33,Default,,0000,0000,0000,,And cause X plus\Nsign X equals 0.
Dialogue: 0,0:30:20.95,0:30:28.01,Default,,0000,0000,0000,,So one of these two brackets,\None or the other, is equal to
Dialogue: 0,0:30:28.01,0:30:33.44,Default,,0000,0000,0000,,0, so cause X minus sign X\Nequals 0 or.
Dialogue: 0,0:30:34.44,0:30:40.68,Default,,0000,0000,0000,,Cause X plus sign\NX equals 0.
Dialogue: 0,0:30:41.52,0:30:46.79,Default,,0000,0000,0000,,Let's develop this one first,\Ncause X minus sign X equals 0
Dialogue: 0,0:30:46.79,0:30:53.37,Default,,0000,0000,0000,,means that they must be the same\Ncause X and sign X are the same.
Dialogue: 0,0:30:54.75,0:31:01.39,Default,,0000,0000,0000,,Divide both sides by Cos X and\Nwe get sign over calls which is
Dialogue: 0,0:31:01.39,0:31:07.22,Default,,0000,0000,0000,,tan. And so Tan X is equal to\Ndividing both sides by Cos X Cos
Dialogue: 0,0:31:07.22,0:31:09.48,Default,,0000,0000,0000,,divided by cause is just one.
Dialogue: 0,0:31:10.41,0:31:17.12,Default,,0000,0000,0000,,Or Look at this\None. Take kozaks away from each
Dialogue: 0,0:31:17.12,0:31:23.97,Default,,0000,0000,0000,,side and we have sine X is equal\Nto minus Cos X. Divide both
Dialogue: 0,0:31:23.97,0:31:29.84,Default,,0000,0000,0000,,sides by cause X sign X over\Ncause exusia gain tanks and
Dialogue: 0,0:31:29.84,0:31:34.72,Default,,0000,0000,0000,,minus cause X divided by Cos X\Nis minus one.
Dialogue: 0,0:31:35.77,0:31:41.38,Default,,0000,0000,0000,,We've broken that down into\Ntwo separate equations.
Dialogue: 0,0:31:42.21,0:31:46.57,Default,,0000,0000,0000,,Let's have a look at how we\Nsolve them. Again, let's assume
Dialogue: 0,0:31:46.57,0:31:50.56,Default,,0000,0000,0000,,that the range of values of X is\Nnot to pie.
Dialogue: 0,0:31:52.16,0:31:56.38,Default,,0000,0000,0000,,And let's sketch the graph of\Ntan in that range.
Dialogue: 0,0:31:57.05,0:32:02.01,Default,,0000,0000,0000,,Sketches don't have to be\Naccurate, just enough to give us
Dialogue: 0,0:32:02.01,0:32:07.87,Default,,0000,0000,0000,,a picture of the symmetry of the\Ncurve to help us solve the
Dialogue: 0,0:32:07.87,0:32:14.57,Default,,0000,0000,0000,,equation. Tan X is one we know\Nthat this is one of those
Dialogue: 0,0:32:14.57,0:32:19.27,Default,,0000,0000,0000,,special angles that its 45\Ndegrees in degrees. But since
Dialogue: 0,0:32:19.27,0:32:25.38,Default,,0000,0000,0000,,we're working in radians, it's X\Nequals π by 4. In other words,
Dialogue: 0,0:32:25.38,0:32:26.79,Default,,0000,0000,0000,,were across here.
Dialogue: 0,0:32:27.69,0:32:34.75,Default,,0000,0000,0000,,One there is π by 4,\Nhalfway between North and pie by
Dialogue: 0,0:32:34.75,0:32:37.69,Default,,0000,0000,0000,,two. So again, this is.
Dialogue: 0,0:32:38.21,0:32:45.04,Default,,0000,0000,0000,,Pie and the one we want is there\Nthat will be halfway between pie
Dialogue: 0,0:32:45.04,0:32:52.85,Default,,0000,0000,0000,,and three Pi by two. So this is\Ngoing to give us the one that is
Dialogue: 0,0:32:52.85,0:32:58.71,Default,,0000,0000,0000,,halfway between pie and three Pi\Nby two, five π by 4.
Dialogue: 0,0:32:59.48,0:33:02.38,Default,,0000,0000,0000,,With this one, where minus one.
Dialogue: 0,0:33:03.25,0:33:08.29,Default,,0000,0000,0000,,So we're down here, meet, sit\Nthere half way between pie by
Dialogue: 0,0:33:08.29,0:33:14.59,Default,,0000,0000,0000,,two and Π, and so X there is\Ngoing to be three π by 4.
Dialogue: 0,0:33:15.16,0:33:20.74,Default,,0000,0000,0000,,And meets the curve again here\Nhalfway between three Pi by two
Dialogue: 0,0:33:20.74,0:33:25.86,Default,,0000,0000,0000,,an 2π. So that's going to be 7\NPi by 4.
Dialogue: 0,0:33:26.67,0:33:31.40,Default,,0000,0000,0000,,So by spotting that we could\Nfactorise this equation, we
Dialogue: 0,0:33:31.40,0:33:36.60,Default,,0000,0000,0000,,didn't need to use the\Nidentity and we came up with
Dialogue: 0,0:33:36.60,0:33:37.55,Default,,0000,0000,0000,,these solutions.
Dialogue: 0,0:33:39.15,0:33:42.82,Default,,0000,0000,0000,,If you want, you can use the
Dialogue: 0,0:33:42.82,0:33:48.17,Default,,0000,0000,0000,,identity. Notice what we get\Nhere are the all the pie by
Dialogue: 0,0:33:48.17,0:33:49.74,Default,,0000,0000,0000,,force if you like.
Dialogue: 0,0:33:50.77,0:33:58.18,Default,,0000,0000,0000,,Only old pie by falls pie by 4\Nfive. PI43 Pi 4 Seven π by 4.
Dialogue: 0,0:34:00.02,0:34:04.24,Default,,0000,0000,0000,,So let's have a look at this\Nequation again, Cos squared X.
Dialogue: 0,0:34:04.85,0:34:08.31,Default,,0000,0000,0000,,Minus sign squared X equals 0.
Dialogue: 0,0:34:09.07,0:34:15.40,Default,,0000,0000,0000,,And we're going to use our basic\Ntrig identity to solve it, so we
Dialogue: 0,0:34:15.40,0:34:21.27,Default,,0000,0000,0000,,know that sine squared X plus\Ncost squared X is equal to 1.
Dialogue: 0,0:34:22.29,0:34:27.45,Default,,0000,0000,0000,,I'm going to replace the sine\Nsquared here, so let's have a
Dialogue: 0,0:34:27.45,0:34:31.75,Default,,0000,0000,0000,,look what is sine squared\Naccording to our identity sign
Dialogue: 0,0:34:31.75,0:34:37.34,Default,,0000,0000,0000,,squared X? If we take away Cos\Nsquared from each side is 1
Dialogue: 0,0:34:37.34,0:34:42.93,Default,,0000,0000,0000,,minus Cos squared X. So I'm\Ngoing to take that, put it in
Dialogue: 0,0:34:42.93,0:34:50.36,Default,,0000,0000,0000,,there. Cos squared X minus,\Nthen a bracket 1 minus
Dialogue: 0,0:34:50.36,0:34:52.69,Default,,0000,0000,0000,,Cos squared X.
Dialogue: 0,0:34:52.70,0:34:57.03,Default,,0000,0000,0000,,And I use the bracket because\NI'm taking away all of this
Dialogue: 0,0:34:57.03,0:35:02.09,Default,,0000,0000,0000,,expression, not just a little\Nbit of it, but all of it. So the
Dialogue: 0,0:35:02.09,0:35:05.70,Default,,0000,0000,0000,,bracket show that now I need to\Nremove the brackets.
Dialogue: 0,0:35:05.80,0:35:13.31,Default,,0000,0000,0000,,Cos squared X minus one and\Nminus minus gives me a plus.
Dialogue: 0,0:35:14.02,0:35:17.16,Default,,0000,0000,0000,,Cos squared X equals 0.
Dialogue: 0,0:35:17.69,0:35:22.92,Default,,0000,0000,0000,,So now I've cost squared plus\Ncost squared. That's two of
Dialogue: 0,0:35:22.92,0:35:25.76,Default,,0000,0000,0000,,them. Two cost squared X equals.
Dialogue: 0,0:35:26.28,0:35:33.21,Default,,0000,0000,0000,,Wall by adding this one to both\Nsides. Now let me divide by two.
Dialogue: 0,0:35:33.81,0:35:37.27,Default,,0000,0000,0000,,Cost squared X is one over 2.
Dialogue: 0,0:35:38.14,0:35:42.73,Default,,0000,0000,0000,,Now at this point I could say\None over 2 and half. That's not
Dialogue: 0,0:35:42.73,0:35:46.67,Default,,0000,0000,0000,,.5 and get my Calculator out\Nbecause I'm going to have to
Dialogue: 0,0:35:46.67,0:35:47.98,Default,,0000,0000,0000,,take a square root.
Dialogue: 0,0:35:48.59,0:35:55.81,Default,,0000,0000,0000,,But I don't want to do that,\Nwhy not? Well, half is a nice
Dialogue: 0,0:35:55.81,0:36:01.90,Default,,0000,0000,0000,,number and I happen to know, for\Ninstance, that sign 30 is 1/2
Dialogue: 0,0:36:01.90,0:36:08.92,Default,,0000,0000,0000,,cost, 60 is 1/2. I also know\Nthat sign of 45 and cause of 45
Dialogue: 0,0:36:08.92,0:36:15.00,Default,,0000,0000,0000,,are both one over the square\Nroot of 2, so there are enough
Dialogue: 0,0:36:15.00,0:36:20.15,Default,,0000,0000,0000,,indications here to suggest to\Nme that there is a nice
Dialogue: 0,0:36:20.15,0:36:24.28,Default,,0000,0000,0000,,relationship. Between the angle\Nand the cosine that I'm going to
Dialogue: 0,0:36:24.28,0:36:29.40,Default,,0000,0000,0000,,get when I take the square root,\Nso I don't want to spoil that
Dialogue: 0,0:36:29.40,0:36:33.80,Default,,0000,0000,0000,,relationship by messing it up\Nwith a lot of decimals through a
Dialogue: 0,0:36:33.80,0:36:38.19,Default,,0000,0000,0000,,Calculator. So let's take that\Nsquare root cause of X is one
Dialogue: 0,0:36:38.19,0:36:40.38,Default,,0000,0000,0000,,over the square root of 2.
Dialogue: 0,0:36:41.33,0:36:47.43,Default,,0000,0000,0000,,But I've taken a square root so\Nthat means not only must I have
Dialogue: 0,0:36:47.43,0:36:50.05,Default,,0000,0000,0000,,plus, but I must have minus.
Dialogue: 0,0:36:50.84,0:36:55.99,Default,,0000,0000,0000,,Now we didn't say at the\Nbeginning what was the range
Dialogue: 0,0:36:55.99,0:37:01.60,Default,,0000,0000,0000,,of values of X, so let's take\Nthe range that we've been
Dialogue: 0,0:37:01.60,0:37:04.88,Default,,0000,0000,0000,,working with, namely between\NNorth and 2π.
Dialogue: 0,0:37:06.08,0:37:10.62,Default,,0000,0000,0000,,So a sketch of the graph just to\Nhelp us see where we are.
Dialogue: 0,0:37:11.70,0:37:14.83,Default,,0000,0000,0000,,Here. Pie by two.
Dialogue: 0,0:37:15.40,0:37:17.44,Default,,0000,0000,0000,,Here pie.
Dialogue: 0,0:37:18.64,0:37:25.08,Default,,0000,0000,0000,,Three Pi by two there and 2π\Nthere and our cosine function
Dialogue: 0,0:37:25.08,0:37:27.77,Default,,0000,0000,0000,,ranges between one and minus
Dialogue: 0,0:37:27.77,0:37:34.43,Default,,0000,0000,0000,,one. We know that the cosine of\NX is one over Route 2. We know
Dialogue: 0,0:37:34.43,0:37:39.61,Default,,0000,0000,0000,,that's 45 degrees or radians π\Nby 4, so we know that we're
Dialogue: 0,0:37:39.61,0:37:46.32,Default,,0000,0000,0000,,here. Arc PY by 4\Nand of course right across there
Dialogue: 0,0:37:46.32,0:37:52.78,Default,,0000,0000,0000,,and again halfway between these\Ntwo, so this bit is telling us
Dialogue: 0,0:37:52.78,0:37:59.23,Default,,0000,0000,0000,,X is π by 4 or its\Npartner is here halfway between
Dialogue: 0,0:37:59.23,0:38:04.61,Default,,0000,0000,0000,,three. Pi by two and 2π Seven\NΠ by 4.
Dialogue: 0,0:38:05.35,0:38:10.60,Default,,0000,0000,0000,,And then 4 - 1 over Route 2,\Nwe're going to be about there.
Dialogue: 0,0:38:11.66,0:38:13.30,Default,,0000,0000,0000,,Go across to the graph.
Dialogue: 0,0:38:14.97,0:38:20.53,Default,,0000,0000,0000,,Up to the X axis and again this\Nby the symmetry of the curve
Dialogue: 0,0:38:20.53,0:38:26.09,Default,,0000,0000,0000,,must be half way between pivi 2\Nand Π, and so that gives us
Dialogue: 0,0:38:26.09,0:38:27.67,Default,,0000,0000,0000,,three π by 4.
Dialogue: 0,0:38:28.56,0:38:33.50,Default,,0000,0000,0000,,And here again, halfway between\Npie and three Pi by two. So
Dialogue: 0,0:38:33.50,0:38:39.27,Default,,0000,0000,0000,,again, that gives us five π by\N4. So in solving this equation a
Dialogue: 0,0:38:39.27,0:38:44.63,Default,,0000,0000,0000,,different way we've got the same\Nset of answers. And again we can
Dialogue: 0,0:38:44.63,0:38:49.57,Default,,0000,0000,0000,,recognize them, because these\Nare the odd pie by force pie by
Dialogue: 0,0:38:49.57,0:38:55.34,Default,,0000,0000,0000,,4, three π by 4, five π by 4,\Nand Seven π by 4.
Dialogue: 0,0:38:56.20,0:39:00.40,Default,,0000,0000,0000,,So whether we do this solution\Nof the equation by.
Dialogue: 0,0:39:01.02,0:39:02.41,Default,,0000,0000,0000,,Using this method.
Dialogue: 0,0:39:03.17,0:39:08.20,Default,,0000,0000,0000,,Using the identity or the method\Nthat we had before where we
Dialogue: 0,0:39:08.20,0:39:12.39,Default,,0000,0000,0000,,factorized it doesn't matter.\NAnd that's true in solving any
Dialogue: 0,0:39:12.39,0:39:16.58,Default,,0000,0000,0000,,of these trig equations. The\Nmethod that you use shouldn't
Dialogue: 0,0:39:16.58,0:39:20.77,Default,,0000,0000,0000,,matter. It should always give\Nthe same set of answers.
Dialogue: 0,0:39:21.68,0:39:28.44,Default,,0000,0000,0000,,But let's just recap where we\Nstarted from these three basic
Dialogue: 0,0:39:28.44,0:39:34.60,Default,,0000,0000,0000,,and fundamental identity's sign\Nsquared X plus cost squared X
Dialogue: 0,0:39:34.60,0:39:37.06,Default,,0000,0000,0000,,is equal to 1.
Dialogue: 0,0:39:38.00,0:39:40.36,Default,,0000,0000,0000,,1
Dialogue: 0,0:39:40.36,0:39:47.32,Default,,0000,0000,0000,,plus. Hot\Nsquared X is equal
Dialogue: 0,0:39:47.32,0:39:50.99,Default,,0000,0000,0000,,tool cosec squared X.
Dialogue: 0,0:39:52.13,0:39:59.64,Default,,0000,0000,0000,,And 1 + 10 squared X\Nis equal to sex squared X.
Dialogue: 0,0:40:00.36,0:40:03.35,Default,,0000,0000,0000,,Those are our three\Nfundamental trigonometric
Dialogue: 0,0:40:03.35,0:40:08.33,Default,,0000,0000,0000,,identities, and they must be\Nlearned. They must be known,
Dialogue: 0,0:40:08.33,0:40:13.31,Default,,0000,0000,0000,,and you must be able to\Nrecognize them whenever you
Dialogue: 0,0:40:13.31,0:40:14.30,Default,,0000,0000,0000,,see them.