Learn to Solve Sin & Cos Equations (Solving Trig Equations) - Part 1 [16] - Free Educational videos for Students in K-12

Learn to Solve Sin & Cos Equations (Solving Trig Equations) - Part 1 [16] - By Math and Science

Transcript


00:00	Well , welcome back . The title of this lesson
00:02	is called solving equations with trig functions . This is
00:05	part one . Essentially what we're doing in this lesson
00:07	is we're learning to solve elementary equations that involve trig
00:11	and metric functions . So just like regular equations ,
00:13	you want to solve for the variable right ? Like
00:15	X squared plus two equals three . You want to
00:17	solve for X in these equations . The variable we
00:20	want to solve for is going to be data right
00:23	? But it gets more complicated because the issue is
00:25	that since the unit circle is a circle and they're
00:28	often multiple angles in a unit circle that have the
00:31	same sine and cosine . Very often in trigonometry ,
00:33	you can actually have multiple angles , multiple answers for
00:37	the the equation that you're trying to solve . So
00:40	what usually happens is when you're given a trig and
00:42	metric equation , it'll say solve this equation and uh
00:47	the answer for data must be between this range between
00:49	zero and 360 degrees or between zero and pi or
00:53	whatever . The problem may give you a range of
00:56	data that is that is acceptable for the answer to
00:59	that equation . So what you need to remember is
01:02	that the previous lessons , we've already talked about this
01:04	, that when you take an arc sine or an
01:06	arc co sign or an arc tangent , the calculator
01:10	or the computer or the whatever device you're using is
01:12	always going to give you the fundamental kind of base
01:16	angle in the base range of those uh , inverse
01:20	trig functions . And so then you might have to
01:23	often take that base angle and look at what the
01:25	problem is asking you for and and add 180 degrees
01:29	or do some other kind of generations to get to
01:32	the actual answer to the problem . Ask you for
01:35	so long story short , inverse sine , inverse coastline
01:37	, inverse tangent . As operations always give you angles
01:40	back in certain ranges , but the problem may say
01:42	the angles that acceptable for answers for these equations are
01:46	in in in a known range . And so you
01:47	may have to to add or subtract or convert .
01:50	This is where you have to start using your brain
01:51	and you can't just like plug things in . You
01:54	have to know what the what the question is asking
01:56	you for . So for instance , let's just say
01:59	let's do a very simple equation . Sine of theta
02:02	is equal to zero , solve for theta . But
02:05	It's also known that Fatah can be only in the
02:08	range between zero and 360 degrees . Notice it's including
02:14	zero but not quite including three and 360 but it's
02:17	bigger than zero , less than 3 60 . So
02:20	that means essentially the whole unit circle what is basically
02:23	telling you is solve that equation . But the angles
02:26	that you ain't right down is your answer can be
02:29	anywhere between zero all the way around 23 60 but
02:33	not including 3 60 because then you'd be back right
02:35	on top of where you started from . So many
02:38	angles at all , even though we know that this
02:40	is a a sign . So when we do the
02:43	arc sine , it's only going to give us angles
02:45	back in a certain range . We have to provide
02:47	essentially all of the angles that satisfied that equation in
02:51	this range . So let's just go through with it
02:54	and you'll see what I'm talking about . All right
02:56	. So I had to be solved this thing .
02:57	We apply an inverse sine to the left and an
03:00	inverse sine to the right . When we apply an
03:02	inverse sine operation to the left , it annihilates the
03:05	sign and all that were left . With this data
03:07	on the right hand side , we have the inverse
03:10	sine of zero . This is what we have to
03:13	do . So we applied the inverse sine to the
03:14	left that dropped it away inverse sine to the right
03:17	. Now we remember that inverse sine from previous lessons
03:22	. We've done this in the last lesson . The
03:24	inverse sine of the arc sine only returns values back
03:27	like from a calculator in this quadrant from negative pi
03:31	over two up to positive pi over two . So
03:35	if you stick numbers in the calculator and hit inverse
03:38	sine inverse sine inverse sine , you'll always get angles
03:40	between negative pi over two and positive pi over two
03:43	. You will never ever get angles anywhere else in
03:45	the circle because that's the kind of the basic range
03:48	that covers all the bases . And so that's the
03:51	only ones you're going to get back . But our
03:52	problem says , give me all angles between zero and
03:55	3 60 that satisfy this . So what you have
03:58	to do is find like the basic angle that works
04:01	and then check and see if there's any additional angles
04:03	around the unit circle that also work . Right ?
04:06	So , let's just go through it . All right
04:08	. What angle is such that the sine of an
04:12	angle is going to give us zero ? All right
04:14	, we do that in a calculator . What do
04:16	we have here ? Well , if you look at
04:17	0° , the sign is the projection onto the y
04:20	axis . So zero degrees . There is no projection
04:23	. So the sign of zero is zero . So
04:25	, we know that feta can be zero degrees or
04:29	zero radiance . And notice that this is the angle
04:32	that will be given back to us in the calculator
04:34	. Go ahead and hit zero in the calculator ,
04:36	inverse sine . And you're gonna get zero degrees or
04:38	radiance , whatever mode you're in . And that is
04:40	right in the middle of this range , negative pi
04:42	over two up to five or two zeros right in
04:44	the middle . So that's why it gave us that
04:45	as an answer . However , are there any other
04:48	angles around the unit circle that also give the sign
04:52	of that angle to be zero ? Well , if
04:53	you look on the other side of the unit circle
04:55	over here at pi , what is the sine of
04:58	pi ? Well , the projection onto the y axis
05:00	over here is also zero . So it's outside the
05:03	range of the fundamental range of the arc sine function
05:06	. The calculator will never give you pies and answer
05:08	. But our equation says give me all angles from
05:11	0 to 3 60 that satisfy this Only in this
05:14	range though . And so we then know that this
05:16	is one of the angles And then we put in
05:19	and down here . And then we say that data
05:21	can then be 180° or pie radiance . Just writing
05:26	it in degrees and radiant . So we get some
05:27	practice with both . So that's what you write down
05:30	. So we're looking for all angles in this range
05:32	where the sign is zero , the sign is zero
05:34	right here , the sign is zero right here .
05:37	If I go around again Notice there's no equal sign
05:40	under 3 60 . So I don't write 360° now
05:43	because it's not included in the range of data that
05:45	it wants . But you I mean obviously you know
05:48	that you can keep spinning around the unit circle ,
05:50	adding 3 60 adding 3 60 over and over and
05:52	you'll get tons of angles . But all of those
05:54	other ones are outside of the range it cares about
05:56	. So that's why I harp so much about what
05:59	the uh kind of the basic range of an arc
06:02	sine our coastline is . That's why I emphasized it
06:05	so much because that's the angle the calculator will give
06:07	you . It's like the base angle . But in
06:10	real equations , you often have a wider range of
06:12	angles that you're hunting for that satisfy the equation .
06:15	So you need to get the base angle first and
06:17	then often you add 1 80 or subtract 1 80
06:19	or something . Notice the first thing we got with
06:21	zero and then we added 1 80 that also satisfies
06:24	the equation . Often you'll be adding or subtracting 1
06:27	80 sometimes other things , but that's a very often
06:30	a very common thing that you will do . So
06:32	that's the solution to problem number one . So ,
06:35	we have two answers and these two answers are equally
06:37	valid When you put 100 and 80 in here signed
06:40	180 0 . When you put zero in here sign
06:42	of zero is also zero . They both work .
06:44	Now let's take a look at the tangent of uh
06:49	I guess I'll put in Princess tangent of Theta is
06:52	equal to zero . Tanja is equal to zero .
06:54	And again , tha can be between zero and 3
06:57	60 . But of course not including 3 60 .
07:00	All right . Now , what we need to do
07:02	to find data is apply the inverse tangent to both
07:04	sides so we can get rid of this and and
07:06	put data by itself . So Fada will be the
07:08	inverse tangent . Arc tangent of zero . Or you
07:12	can write it as tangent with little negative one there
07:15	. All right . So , essentially what you're trying
07:17	to do is you're trying to take the arc tangent
07:20	of zero . Now , you know that tangent or
07:23	arc tangent is going to only as a function .
07:25	It's only going to give you values back in the
07:27	right hand side of the plane . From negative pi
07:30	over two up to pi over two . That's what
07:31	we learned a long time ago . That's the basic
07:34	answer . And you may have to look for additional
07:36	answers in this range that satisfy the equation . So
07:39	you can think of it like this but I actually
07:41	prefer especially for tangent to come down here and say
07:44	what we're really hunting for is the sign of some
07:47	angle over the co sign of some angle and the
07:50	and the rain and the angle is going to be
07:51	in this range here um is gonna be equal to
07:54	zero . Now , if you have sine over cosine
07:56	equals zero . What you're trying to figure out is
07:59	what angles make it such that when you do the
08:01	division you get zero . So what this really means
08:03	that since you want to zero here you want sign
08:06	of the angle to be zero . Because if the
08:09	numerator goes to zero you have a zero over anything
08:12	you want is going to give you zero even infinity
08:14	. So it doesn't really matter . Well it does
08:17	matter but I'm just trying to tell you that to
08:19	figure out the angles that make it go to zero
08:21	. All you have to do is find out where
08:22	the numerator goes to zero . Where is sign of
08:25	zero equals zero . Where a sign of zero equals
08:28	zero . The angle that you're looking for is zero
08:31	degrees or zero radiance . So if you go put
08:35	zero in a calculator and do the inverse to do
08:38	the tangent of zero , you're gonna get zero .
08:41	Um and it's also in the proper range . Alright
08:44	, so here we are 0°. . But are there
08:47	any other angles in this full circle range that make
08:51	the tangent go to zero basically ? Are there any
08:53	other angles that make the sign go to zero ?
08:55	It's the same questions before we go over here to
08:58	pie . And over here at Pi Sine of Pi
09:01	is also going to be zero . So we also
09:03	say and FAHA can be pie or 0° I guess
09:09	I wrote it backwards , but you get the idea
09:11	. So actually in this case we have two different
09:13	equations and we get these exact same answers . And
09:16	the reason that we get the exact same answers because
09:18	the first equation was the sign . Tell me where
09:21	the angles where the sign of that angle is zero
09:23	. The second equation is tell me the angles where
09:26	the tangent is zero . But when you think about
09:28	it , that's the same angles where the sign is
09:30	also zero . So it gives you the exact same
09:32	answer , even though it's a different equation . And
09:34	we have found all the angles in this range first
09:37	by finding the fundamental angle in this range and then
09:40	, you know , in this case adding 180 to
09:42	figure out additional ones . But I don't want you
09:45	to think , Oh I'll just add 180 every time
09:47	because it's not going to be the case , you
09:48	have to think through it , right ? So let's
09:51	take a look at the next one . Let's say
09:53	we have the tangent of some angle theta and that's
09:57	equal to one . And in this case the range
10:00	of data for this equation can be between zero and
10:04	3 , 16 . And for this lesson , all
10:06	of the ranges of data will be here . But
10:09	just know that when you solve other equations down the
10:11	line , more complicated equations , the range of data
10:14	that you're told to to bind yourself by or to
10:17	look for will be different . Okay , it maybe
10:20	even between zero and 90 or zero and 100 and
10:22	80 it'll lock it down in different ways . Here
10:24	, we're just starting . So we just choose the
10:27	entire unit circle . Tell me all angles in the
10:29	unit circle where the tangent of the angle is one
10:32	. Okay , so you do the inverse tangent uh
10:35	to to um to both sides . And what you
10:38	basically figure out is that the uh data is going
10:42	to be the inverse tangent of one , right ?
10:47	The inverse tangent of the left , inverse tangent of
10:49	the right . And so you go over here and
10:50	say , well again , where is it going to
10:52	give me angles back ? The tangent function gives me
10:55	angles over here . All right . So you can
10:59	think of it that way . You know I mean
11:01	it's fine to say inverse tangent of one . But
11:03	really , even though I wrote this here , really
11:05	, I like to come down here and say ,
11:06	well the tangent is the sine of the angle divided
11:09	by the co sine of the angle . And basically
11:11	I'm trying to figure out all locations around the unit
11:14	circle where the sign of the coastline is one .
11:16	So when you think through it though , the tangent
11:20	is sine over the co sign . So the only
11:22	way that can be one is if the sign and
11:24	the co signer equal in magnitude right equal . And
11:27	that's only at 45 degrees because you know , at
11:29	45 degrees . The sign is a squared of 2/2
11:32	. And the co sign also the squared of two
11:34	over to so . And that is also in the
11:36	correct quadrant uh for the tangent up there or the
11:39	correct , You know uh 45° up here is in
11:42	the shaded region . So we then say that data
11:45	can be 45° or pi over four , just converting
11:50	its radiance there . And the reason that , you
11:53	know that I'll just kind of write it off to
11:54	the side . It's because the sine of pi over
11:59	four over the co sign of pi over four is
12:04	square root 2/2 over square root 2/2 . And that
12:08	equals one . So that angle is the only angle
12:11	that gives you a positive uh tangent of one like
12:14	this and it's in the correct quadrant . Now the
12:18	quadrant , the actual range of angles that we need
12:20	to hunt for is the entire unit circle . So
12:23	are there any other um angles where the tangent will
12:28	be a positive one like this ? So you start
12:31	thinking about it right ? Let's go over to our
12:32	unit circle and actually look . So we basically said
12:36	is the base angle for the solution to that equation
12:39	? Is that pirate before ? Because Sine over cosine
12:42	is one that's the tangent right Now if we go
12:44	over to this quadrant here , if we try to
12:46	divide sign divided by co sign , we're still gonna
12:49	get a one but it'll be a negative one because
12:51	we divide here . If we go to this quadrant
12:54	over here we'll have a sign divided by a coastline
12:56	. But again it'll be negative one . So it
12:58	can't be this quadrant and it can't be this quadrant
13:00	. But if we spin all the way around totally
13:02	diagonal . So this one over here at 225 degrees
13:05	, the sign is negative square 2/2 . The coastline
13:09	is also -2 or two because the projections are on
13:11	the negative access . But when you divide them you
13:13	still get a positive one . So basically by taking
13:17	this and again adding 180° and going totally opposite .
13:21	We get to another quadrant where the sign and the
13:24	co signer negative . But the tangent is still a
13:26	positive number so that we can say . And feta
13:30	can be 225 degrees or uh five pi over for
13:37	me . Just check myself 225 degrees or 554 And
13:40	you circle both of those . So we're trying to
13:42	find all angles within 360 degree circle that work 45°.
13:46	. Works right here . It's kind of like put
13:50	this out here , pi over four works and then
13:52	if you spend all the way around over here ,
13:55	five pi over four or 225 degrees also works in
13:59	quadrant . One sign over co sign gives you positive
14:03	one . In quadrant three . Sign over coastline also
14:06	gives you positive one because that sign and the coastline
14:09	are both negative here . In this quadrant , the
14:11	tangents negative and in this quadrant the tangents also negative
14:14	because you have opposite signs on the sign and the
14:16	coastline . So with Tanja , you really have to
14:18	think about what quadrant you're in to get the correct
14:20	answer . All right now , the last problem we
14:24	have . I don't really want to say it's tricky
14:26	, but it does require us to think just a
14:28	little bit . All right , so let's solve the
14:32	equation sign of feta Is equal to negative one .
14:36	And let's do it in the same range . Data
14:38	has to be bigger than or equal to zero in
14:41	less than 360°. . So what we'll do is the
14:46	same thing . We always do . Well come down
14:48	here and we'll say imply the inverse sine to the
14:50	left and the inverse sine to the right . So
14:52	we'll have data is inverse sine of -1 . And
14:57	we know that the inverse sine is going to prove
15:01	in a calculator . Anyway , it's gonna give us
15:03	angles back that are gonna be in the right hand
15:05	plane like this . From negative pi over two up
15:08	to positive pi over two . Okay , so let's
15:11	go and take a look at this and see what
15:12	happens essentially . What we're asking ourselves is what angle
15:16	in that range is such that the sign of that
15:19	angle is negative one . So , where can it
15:21	be ? What is the sign of zero sign of
15:23	zero ? Zero ? Because there's no projection on the
15:25	y axis . What's the sine of pi over two
15:27	? Up here . The entire thing lies on the
15:30	lines of the sign is one , but we want
15:32	negative one . So if we go down here then
15:34	we see , okay , down here at this angle
15:36	, the sign is a negative one as well .
15:41	Okay , so what we right here is to say
15:43	, well the base angle of what the calculator would
15:45	give us go and put it in the calculator ,
15:47	hit negative one , hit the inverse sine button and
15:49	see what it gives you . It's going to tell
15:50	you that the angle is negative pi over two ,
15:53	which is exactly in the range of what we said
15:55	, negative pi over two , up to pie or
15:57	two for the sign . And it's giving me negative
15:59	pi over two because here the projection is exactly on
16:01	the negative one part of the y axis . So
16:04	this is the base angle , right negative pi over
16:06	two . However , look at the answers it wants
16:09	us to give us it says solve this equation but
16:12	give me angles back that exists from zero up to
16:14	360 degrees . Let me ask you 0 to 3
16:18	60 . 0 to 3 60 is negative pi over
16:21	to between zero and 3 60 or zero and two
16:23	pi . If you want to think about ratings .
16:25	Is it in the range ? I mean we know
16:26	that it's down here and we can find another angle
16:29	that has the same label . But literally negative pi
16:32	over two is not in the range because negative pi
16:34	over two is actually even less than zero . So
16:37	the this is an example of when the bass range
16:39	that a calculator gives you back is not in the
16:42	range that they tell you the angle can be for
16:45	the answer to that equation . So even though the
16:47	calculator gives you that angle , you cannot circle ,
16:49	that angle is the answer . This is a great
16:52	way to get a question wrong without really because you're
16:54	not thinking about it , it gave you an answer
16:57	outside of the range of what you're allowed to get
17:00	. So how do you figure that out ? So
17:01	then you say well , ok , I can't put
17:03	pi over two down there . So what angle is
17:05	in that range ? I know it's down here .
17:07	So you go , well this is pi over two
17:09	, I'm sorry , this is pi over two then
17:11	to pi over two , this is three pi over
17:13	two down here . So what you then say is
17:14	data is really equal to three pi over two or
17:19	um 270 degrees . This is the angle that you
17:23	circle because three pi over two or 270 degrees is
17:26	in the range , noticed 2 70 is right here
17:29	. So if you want to check yourself , go
17:32	down here to to 73 pi over two , the
17:33	sine is negative one . So you can verify that
17:36	. It's correct . I realize and I recognize that
17:38	this gets confusing because I've been telling you that the
17:41	arc sine only gives you angles back in a certain
17:43	range and now I'm telling you to reject the answer
17:46	that it gives you and give me another answer .
17:48	Well that's that's the way the game is played .
17:50	The calculator . As you get farther in math ,
17:52	you're gonna realize just using a calculator is not going
17:54	to help you so much . Because the answer is
17:56	a calculator gives , you can be outside of the
17:59	range of what the problem is telling you to give
18:01	you . The problem is saying only give me angles
18:04	that make this equation work between zero and 360 degrees
18:08	. Not including 360 . Don't give me any angles
18:10	outside of that . So you stick this in there
18:12	. You do the inverse sine or you think about
18:15	the where it is in the circle and say ,
18:16	okay , that's a negative pi over two . So
18:18	you circle pi over two and it's wrong . It's
18:20	because negative private use not in this rain . So
18:22	you have to think . You have to you can't
18:24	just do you have to think , Okay , the
18:26	angle is here . It's labeled as negative pi over
18:28	two because that's the base angle of what the calculator
18:30	will give me . But I need to express this
18:32	location in this range . Okay . So I got
18:35	to go in a positive sense poverty to private 23
18:37	prior to this is the same location , same angle
18:40	now it's in the range and in this case I
18:43	only have one angle in this range of a complete
18:47	circle . That gives me the sign equal to negative
18:50	one and all the other problems that had two angles
18:52	that worked and there were 180° apart . Two angles
18:55	work 180 degrees apart . Two angles work 180 degrees
18:58	apart . Here I give you a problem where you
19:00	do not have to angles 180 degrees apart . You
19:02	only have one angle and also you have to reject
19:05	your base angle anyway and count in the positive sense
19:07	to get there . So that's why I can't tell
19:10	you . Oh yeah . Just get the answers and
19:11	add 1 80 all the time . You can't you
19:13	can't do that . You have to know what the
19:15	problem is asking and think logically about what's required .
19:18	This problem only has one solution in this range .
19:21	Okay , so make sure you can solve these yourself
19:24	. They're all good little problems . They're not super
19:26	hard , but they do require you to think .
19:27	Follow me on to the next lesson . We're gonna
19:29	get more practice with solving equations that involve trigonometry ,
19:33	trig and metric functions .

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