## www.mathcentre.ac.uk/.../5.5Solving%20Trigonometric%20Equations.mp4

• 0:01 - 0:06
In this video, we're going to be
solving whole collection of
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trigonometric equations now be
cause it's the technique of
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solving the equation and in
ensuring that we get enough
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solutions, that's important and
not actually looking up the
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angle. All of these are designed
around certain special angles,
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so I'm just going to list at the
very beginning here the special
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angles and their sines, cosines,
and tangents that are going to
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form. The basis of what
we're doing.
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So the special angles that we're
going to have a look at our
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zero. 30 4560
and 90 there in degrees.
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then there's zero.
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Pie by 6.
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Pie by 4.
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Pie by three.
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And Π by 2.
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Trig ratios we're going to be
looking at are the sign.
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The cosine. On the tangent
of each of these.
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Sign of 0 is 0.
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The sign of 30 is 1/2.
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Sign of 45 is one over Route 2.
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The sign of 60 is Route 3 over 2
and the sign of 90 is one.
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Cosine of 0 is one.
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Cosine of 30 is Route 3 over 2
cosine of 45 is one over Route
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2, the cosine of 60 is 1/2, and
the cosine of 90 is 0.
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The town of 0 is 0 the
town of 30 is one over
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Route 3 that Anna 45 is
110 of 60 is Route 3 and
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the town of 90 degrees is
infinite, it's undefined.
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It's these that we're going to
be looking at and working with.
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Let's look at our first
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equation then. We're going to
begin with some very simple
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ones. So we take sign of
X is equal to nought .5. Now
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invariably when we get an
equation we get a range of
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values along with it.
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So in this case will take X is
between North and 360. So what
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we're looking for is all the
values of X.
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Husain gives us N
.5.
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Let's sketch a graph of
sine X over this range.
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And sign looks like that
with 90.
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180
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270 and 360 and ranging between
one 4 sign 90 and minus
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one for the sign of 270.
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Sign of X is nought .5. So
we go there.
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And there.
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So there's our first angle, and
there's our second angle.
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We know the first one is 30
degrees because sign of 30 is
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1/2, so our first angle is 30
degrees. This curve is symmetric
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and so because were 30 degrees
in from there, this one's got to
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be 30 degrees back from there.
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That would make it
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150. There are no more answers
because within this range as we
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go along this line.
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It doesn't cross the curve at
any other points.
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Let's have a look
at a cosine cause
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of X is minus
nought .5 and the
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range for this X
between North and 360.
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So again, let's have a look at a
graph of the function.
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Involved in the equation,
the cosine graph.
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Looks like that. One and
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minus one.
This is 90.
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180
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270 and then here
at the end, 360.
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Minus 9.5.
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Gain across there at minus
9.5 and down to their and
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down to their.
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Now the one thing we do know is
that the cause of 60 is plus N
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.5, and so that's there. So we
know there is 60. Now again,
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this curve is symmetric, so if
that one is 30 back that way
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this one must be 30 further on.
So I'll first angle must be 120
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degrees. This one's got to be in
a similar position as this bit
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of the curve is again symmetric.
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So that's 270 and we need
to come back 30 degrees, so
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that's 240. Now we're going to
have a look at an example where
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we've got what we call on
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just being cause of X or sign of
X, it's going to be something
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like sign of 2X or cause of
three X. So let's begin with
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sign of. 2X is equal
to Route 3 over 2
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and again will take X
to be between North and
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360.
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Now we've got 2X here.
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So if we've got 2X and X
is between Norton 360, then the
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total range that we're going to
be looking at is not to 722.
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X is going to come between 0
and 720, and the sign function
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is periodic. It repeats itself
every 360 degrees, so I'm going
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to need 2 copies of the sine
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curve. As the first one going up
to 360 and now I need a second
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copy there going on till.
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720
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OK, so sign 2 X equals root, 3
over 2, but we know that the
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sign of 60 is Route 3 over 2. So
if we put in Route 3 over 2 it's
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there, then it's going to be
these along here as well. So
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what have we got? Well, the
first one here we know is 60.
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This point we know is 180 so
that one's got to be the same
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distance. Back in due to the
symmetry 120, so we do know that
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2X will be 60 or 120, but we
also now we've got these other
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points on here, so let's just
count on where we are. There's
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the 1st loop of the sign
function, the first copy, its
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periodic and repeats itself
again. So now we need to know
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where are these well.
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This is an exact copy of
that, so this must be 60
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further on. In other words, at
420, and this must be another
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120 further on. In other words,
at 480. So we've got two
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that we actually want, not
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2X. So this is 3060.
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210 and finally
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240. Let's have a
look at that with a tangent
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function. This time tan or three
X is equal to.
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Minus one and will
take X to be
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between North and 180.
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So we draw a graph
of the tangent function.
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So we go up.
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We've got that there. That's 90.
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This is 180 and this is 270
now. It's 3X. X is between
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Norton 180, so 3X can be between
North and 3 * 180 which is
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540. So I need to get copies
of this using the periodicity of
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the tangent function right up to
540. So let's put in some more.
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That's 360. On
there.
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That's 450.
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This one here will be
540 and that's as near
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or as far as we
need to go. Tanner 3X
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is minus one, so here's
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minus one. And we go across here
picking off all the ones that we
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need. So we've got one there.
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There there. These are our
values, so 3X is equal 12.
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Now we know that the angle
whose tangent is one is 45,
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which is there. So again this
and this are symmetric bits of
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curve, so this must be 45
further on. In other words 130.
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5.
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This one here has got to be
45 further on, so that will be
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315. This one here has got
to be 45 further on, so that
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will be 495, but it's X that
we want not 3X, so let's divide
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throughout by three, so freezing
to that is 45 threes into that
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is 105 and threes into that is
165. Those are our three answers
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for that one 45 degrees.
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105 degrees under
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165. Let's take cause of
X over 2 this time. So
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instead of multiplying by two or
by three, were now dividing by
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two. Let's see what difference
this might make equals minus 1/2
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and will take X to be
between North and 360. So let's
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draw the graph.
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All calls X between North
and 360, so there we've
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got it 360 there.
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180 there, we've got 90 and 270
there in their minus. 1/2 now
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that's going to be.
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Their cross and then these are
the ones that we are after.
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So let's work with that. X over
2 is equal tool. Now where are
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we? Well, we know that the angle
whose cosine is 1/2 is in fact
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60 degrees, which is here 30 in
from there. So that must be 30
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further on. In other words, 120
and this one must be 30 back. In
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other words, 240. So now we
multiply it by.
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Two, we get 240 and 480, but
of course this one is outside
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the given range. The range is
not to 360, so we do not
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need that answer, just want the
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240. Now we've been working with
a range of North 360, or in one
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case not to 180, so let's change
the range now so it's a
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symmetric range in the Y axis,
so the range is now going to run
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from minus 180 to plus 180
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degrees. So we'll begin with
sign of X equals 1X is to
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be between 180 degrees but
greater than minus 180 degrees.
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Let's sketch the graph of sign
in that range. So we want to
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complete copy of it. It's going
to look like that.
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Now we know that the angle
who sign is one is 90
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degrees and so we know
that's one there and that's
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90 there and we can see that
there is only the one
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solution it meets the curve
once and once only, so
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that's 90 degrees.
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Once and once only, that is
within the defined range. Let's
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take another one.
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So now we use a multiple
angle cause 2 X equals 1/2
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and will take X to be
between minus 180 degrees and
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plus 180 degrees. So let's
sketch the graph. Let's remember
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that if X is between minus
180 and plus one 80, then
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2X will be between minus 360.
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And plus 360.
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So what we need to do is use the
periodicity of the cosine
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function to sketch it.
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In the range. So there's
the knocked 360 bit and
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then we want.
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To minus 360. So I just label up
the points. Here is 90.
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180
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Two 7360 and then back
this way minus 90 -
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180. Minus 270 and
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minus 360. Now cause
2X is 1/2, so here's a half.
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Membrane that this goes between
plus one and minus one and if we
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draw a line across to see where
it meets the curve.
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Then we can see it meets it in
four places. There, there there
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and there we know that the angle
where it meets here is 60
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degrees. So our first value is 2
X equals 60 degrees.
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By symmetry, this one back here
has got to be minus 60.
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again, symmetry says that we are
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60 from here, so we've got to
be 60 back from there, so this
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must be 300 and our symmetry of
the curve says that this one
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must be minus 300, and so we
have X is 30 degrees minus 30
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degrees, 150 degrees and minus
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150 degrees. Working with
the tangent function tan, two
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X equals Route 3 and
again will place X between
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180 degrees and minus 180
degrees. We want to sketch
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the function for tangent and
we want to be aware
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that we've got 2X.
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So since X is between minus 118
+ 182, X is got to be between
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minus 360 and plus 360.
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So if we take the bit between.
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North And 360.
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Which is that bit of the curve
we need a copy of that between
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minus 360 and 0 because again
the tangent function is
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periodic, so we need this bit.
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That
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And we need that and it's Mark
off this axis so we know where
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we are. This is 90.
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180
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270 and 360. So this
must be minus 90 -
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180 - 270 and minus
360.
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Now 2X is Route 3, the angle
whose tangent is Route 3. We
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know is 60, so we go across here
at Route 3 and we meet the curve
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there and there.
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And we come back this way. We
meet it there and we meet there.
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So our answers are down here.
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Working with this one, first we
know that that is 60, so 2X is
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equal to 60 and so that that one
is 60 degrees on from that
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point. Symmetry says there for
this one is also 60 degrees on
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from there. In other words, it's
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240. Let's work our way
backwards. This one must be 60
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degrees on from minus 180, so it
must be at minus 120. This one
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is 60 degrees on.
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From minus 360 and so therefore
it must be minus 300.
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And so if we divide throughout
by two, we have 31120 -
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60 and minus 150 degrees. We
want to put degree signs on
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all of these, so there are
four solutions there.
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Trick equations often come up as
a result of having expressions
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or other equations which are
rather more complicated than
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that and depends upon
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identity's. So I'm going to
have a look at a couple
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of equations. These equations
both dependa pawn two identity's
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that is expressions involving
trig functions that are true for
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all values of X.
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So the first one is sine
squared of X plus cost
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squared of X is one. This is
true for all values of X.
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The second one we derive from
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this one. How we derive it
doesn't matter at the moment,
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but what it tells us is that sex
squared X is equal to 1 + 10
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squared X. So these are the two
identity's that I'm going to be
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using. Sine squared X plus cost
squared X is one and sex squared
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of X is 1 + 10 squared of
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X OK. So how do we go
about using one of those to do
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an equation like this? Cos
• 22:36 - 22:42
squared X? Plus cause
of X is equal
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to sine squared of
X&X is between 180
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and 0 degrees.
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Well.
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We've got a cost squared, A
cause and a sine squared.
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If we were to use our identity
sine squared plus cost squared
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is one to replace the sine
squared. Here I'd have a
• 23:09 - 23:15
quadratic in terms of Cos X, and
if I got a quadratic then I know
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I can solve it either by
Factorizing or by using the
• 23:20 - 23:25
formula. So let me write down
sign squared X plus cost
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squared. X is equal to 1, from
which we can see.
• 23:29 - 23:37
Sine squared X is equal to
1 minus Cos squared of X,
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so I can take this and
plug it into their. So my
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equation now becomes cost
squared X Plus X is equal
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to 1 minus Cos squared X.
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I want to get this as
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term. Constant term equals 0, so
I begin by adding cost squared
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to both sides.
• 24:10 - 24:16
So adding on a cost squared
there makes 2 Cos squared X plus
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cause X equals 1. 'cause I added
cost square to get rid of that
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one. Now I need to take one away
from both sides to cost squared
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X Plus X minus one equals 0.
• 24:35 - 24:39
Now this is just a quadratic
equation, so the first question
• 24:39 - 24:44
I've got to ask is does it
factorize? So let's see if we
• 24:44 - 24:46
can get it to factorize.
• 24:47 - 24:51
I'll put two calls X in there
and cause X in there because
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that 2 cause X times that cause
X gives Me 2 cost squared and I
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put a one under one there 'cause
one times by one gives me one
• 25:02 - 25:07
and now I know to get a minus
sign. One's got to be minus and
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one's got to be plus now I want
plus cause X so if I make this
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one plus I'll have two cause X
times by one.
• 25:17 - 25:22
Is to cause X if I make this one
minus I'll have minus Cos X from
• 25:22 - 25:26
there. Taking those two
together, +2 cause X minus Cos X
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is going to give me the plus
Kozaks in there, so that equals
• 25:31 - 25:36
0. Now, if not equal 0, I'm
multiplying 2 numbers together.
• 25:36 - 25:42
This one 2 cause X minus one and
this one cause X plus one, so
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one of them or both of them have
got to be equal to 0.
• 25:49 - 25:55
So 2 calls X minus
one is 0.
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All cause of X Plus One is
0, so this one tells me that
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cause of X is equal to 1/2.
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And this one tells Maine that
cause of X is equal to minus
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one, and both of these are
possibilities. So I've got to
• 26:18 - 26:23
solve both equations to get the
total solution to the original
• 26:23 - 26:28
equation. So let's begin with
this cause of X is equal to 1/2.
• 26:29 - 26:36
And if you remember the range
of values was nought to 180
• 26:36 - 26:42
degrees, so let me sketch
cause of X between North and
• 26:42 - 26:47
180 degrees, and it looks
like that zero 9180.
• 26:48 - 26:53
We go across there at half and
come down there and there is
• 26:53 - 26:59
only one answer in the range, so
that's X is equal to 60 degrees.
• 27:00 - 27:07
But this one again let's sketch
cause of X between North and
• 27:07 - 27:14
180. There and there between
minus one and plus one and we
• 27:14 - 27:20
want cause of X equal to minus
one just at one point there and
• 27:20 - 27:26
so therefore X is equal to 180
degrees. So those are our two
• 27:26 - 27:30
• 27:30 - 27:34
So it's now have a look at
• 27:34 - 27:41
three. 10 squared X is
equal to two sex squared X
• 27:41 - 27:45
Plus One and this time will
• 27:45 - 27:51
take X. To be between North and
180 degrees. Now, the identity
• 27:51 - 27:56
that we want is obviously the
one, the second one of the two
• 27:56 - 28:01
that we had before. In other
words, the one that tells us
• 28:01 - 28:08
that sex squared X is equal to 1
+ 10 squared X and we want to be
• 28:08 - 28:14
able to take this 1 + 10 squared
and put it into their. So we've
• 28:14 - 28:21
got 3. 10 squared
X is equal to 2
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* 1 + 10 squared
X Plus one. Multiply out
• 28:29 - 28:37
this bracket. 310 squared X
is 2 + 210 squared
• 28:37 - 28:39
X plus one.
• 28:40 - 28:45
We can combine the two and the
one that will give us 3.
• 28:45 - 28:51
And we can take the 210 squared
X away from the three times
• 28:51 - 28:57
squared X there. That will give
us 10 squared X. Now we take the
• 28:57 - 29:04
square root of both sides so we
have 10X is equal to plus Route
• 29:04 - 29:06
3 or minus Route 3.
• 29:08 - 29:11
And we need to look at each of
• 29:11 - 29:15
these separately. So.
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Time X equals Route
• 29:19 - 29:26
3. And Tan X
equals minus Route 3.
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180, so let's have a sketch of
• 29:34 - 29:40
the graph of tan between those
values, so there is 90.
• 29:42 - 29:49
And there is 180 the angle whose
tangent is Route 3, we know.
• 29:50 - 29:57
Is there at 60 so we
know that X is equal to
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60 degrees? Here we've
got minus Route 3, so
• 30:02 - 30:04
again, little sketch.
• 30:05 - 30:10
Between North and 180 range over
which were working here, we've
• 30:10 - 30:16
got minus Route 3 go across
there and down to their and
• 30:16 - 30:22
symmetry says it's got to be the
same as this one. Over here it's
• 30:22 - 30:29
got to be the same either side.
So in fact if that was 60 there
• 30:29 - 30:35
this must be 120 here, so X is
equal to 120 degrees.
• 30:35 - 30:40
So far we've been working in
degrees, but it makes little
• 30:40 - 30:43
difference if we're actually
• 30:43 - 30:49
just have a look at one or two
examples where in fact the range
• 30:49 - 30:55
of values that we've got is in
radians. So if we take Tan, X is
• 30:55 - 31:01
minus one and we take X to be
between plus or minus pie.
• 31:01 - 31:06
Another way of looking at
that would be if we were in
• 31:06 - 31:11
degrees. It will be between
plus and minus 180. Let's
• 31:11 - 31:14
sketch the graph of tangent
within that range.
• 31:16 - 31:17
Up to there.
• 31:18 - 31:21
That's π by 2.
• 31:23 - 31:26
Up to their which is π.
• 31:26 - 31:33
Minus Π
by 2.
• 31:36 - 31:40
Their minus
• 31:40 - 31:45
pie. Ton of X is
minus one, so somewhere
• 31:45 - 31:49
across here it's going to
meet the curve and we can
• 31:49 - 31:51
see that means it here.
• 31:52 - 31:56
And here giving us these
solutions at these points. Well,
• 31:56 - 32:02
we know that the angle whose
tangent is plus one is π by 4.
• 32:02 - 32:09
So this must be pie by 4 further
on, and so we have X is equal to
• 32:09 - 32:16
pie by 2 + π by 4. That will be
3/4 of Π or three π by 4, and
• 32:16 - 32:18
this one here must be.
• 32:19 - 32:25
Minus Π by 4 back there, so
minus π by 4.
• 32:26 - 32:30
Let's take one with
a multiple angle.
• 32:32 - 32:39
So we'll have a look cause
of two X is equal to Route
• 32:39 - 32:41
3 over 2.
• 32:42 - 32:48
I will take
X between North
• 32:48 - 32:55
and 2π. Now if
X is between North and 2π, and
• 32:55 - 32:56
we've got 2X.
• 32:57 - 33:01
And that means that 2X can be
between North and four π.
• 33:02 - 33:07
So again, we've got to make use
of the periodicity.
• 33:08 - 33:12
Of the graph of cosine to get a
second copy of it.
• 33:14 - 33:21
So there's the first copy
between North and 2π, and now we
• 33:21 - 33:27
want a second copy that goes
from 2π up till four π.
• 33:28 - 33:33
We can mark these off that one
will be pie by two.
• 33:33 - 33:34
Pie.
• 33:35 - 33:37
Three π by 2.
• 33:38 - 33:46
This one will be 5 Pi by
two. This one three Pi and this
• 33:46 - 33:49
one Seven π by 2.
• 33:50 - 33:55
So where are we with this cost?
2 X equals. Well, in fact we
• 33:55 - 34:01
know cost to access Route 3 over
2. We know that the angle that
• 34:01 - 34:07
gives us the cosine that is
Route 3 over 2 is π by 6. So
• 34:07 - 34:13
I'll first one is π Phi six,
root 3 over 2. Up here we go
• 34:13 - 34:18
across we meet the curve we come
down. We know that this one here
• 34:18 - 34:20
is π by 6.
• 34:20 - 34:26
Let's keep going across the
curves and see where we come to,
• 34:26 - 34:32
what we come to one here which
is π by 6 short of 2π. So
• 34:32 - 34:40
let me write it down as 2π -
Π by 6, and then again we come
• 34:40 - 34:45
to one here. Symmetry suggests
it should be pie by 6 further
• 34:45 - 34:51
on, so that's 2π + π by 6,
and then this one here.
• 34:51 - 34:58
Is symmetry would suggest his
pie by 6 short of four Pi,
• 34:58 - 35:05
so four π - π by 6.
So let's do that arithmetic 2X
• 35:05 - 35:07
is π by 6.
• 35:08 - 35:13
Now, how many sixths are there
in two? Well, the answer. Is
• 35:13 - 35:19
there a 12 of them and we're
going to take one of them away,
• 35:19 - 35:26
so that's eleven π by 6. We're
going to now add a 6th on, so
• 35:26 - 35:28
that's 13 Pi by 6.
• 35:30 - 35:37
How many 6th are there in four
or there are 24 of them? We're
• 35:37 - 35:44
going to take one away, so
that's 23. Pi over 6. Now we
• 35:44 - 35:51
want X, so we divide each of
these by 2π by 1211 Pi by
• 35:51 - 35:59
12:13, pie by 12, and 20, three
π by 12, and there are our
• 35:59 - 36:06
four solutions. Let's have a
look at one where we've got the
• 36:06 - 36:13
X divided by two rather than
multiplied by two. So the sign
• 36:13 - 36:18
of X over 2 is minus Route
3 over 2.
• 36:19 - 36:26
And let's take X to be
between pie and minus π. So
• 36:26 - 36:33
will sketch the graph of sign
between those limited, so it's
• 36:33 - 36:40
there. And their π
zero and minus pie.
• 36:41 - 36:47
Where looking for minus three
over 2. Now the one thing we do
• 36:47 - 36:53
know is that the angle who sign
is 3 over 2 is π by 3.
• 36:54 - 36:59
But we want minus Route 3 over
2, so that's down there.
• 37:00 - 37:02
We go across.
• 37:02 - 37:05
And we meet the curve these two
• 37:05 - 37:09
points. Now this curve is
symmetric with this one.
• 37:09 - 37:12
So if we know that.
• 37:13 - 37:15
Plus Route 3 over 2.
• 37:15 - 37:21
This one was Pi by three. Then
we know that this one must be
• 37:21 - 37:23
minus π by 3.
• 37:23 - 37:31
This one is π by three back, so
it's at 2π by three, so this one
• 37:31 - 37:38
must be minus 2π by three, and
so we have X over 2 is equal
• 37:38 - 37:45
to minus 2π by three and minus,
π by three, but it's X that we
• 37:45 - 37:52
want, so we multiply up X equals
minus four Pi by three and minus
• 37:52 - 37:53
2π by 3.
• 37:54 - 38:00
Let's just check on these
values. How do they fit with the
• 38:00 - 38:06
given range? Well, this 1 - 2π
by three is in that given range.
• 38:07 - 38:11
This one is outside, so we don't
want that one.
• 38:12 - 38:19
A final example here, working
with the idea again of using
• 38:19 - 38:25
those identities and will take 2
cost squared X.
• 38:25 - 38:31
Plus sign X is
equal to 1.
• 38:32 - 38:38
And we'll take X between
North and 2π.
• 38:39 - 38:43
We've got causes and signs,
so the identity that we're
• 38:43 - 38:48
going to want to help us
will be sine squared plus
• 38:48 - 38:50
cost. Squared X equals 1.
• 38:51 - 38:53
Cost squared here.
• 38:54 - 39:00
Cost squared here. Let's use
this identity to tell us that
• 39:00 - 39:06
cost squared X is equal to 1
minus sign squared X and make
• 39:06 - 39:09
the replacement up here for cost
• 39:09 - 39:14
squared. Because that as we will
see when we do it.
• 39:15 - 39:23
so it's multiply this out 2 -
• 39:23 - 39:31
2 sine squared X plus sign X
is equal to 1 and I want
• 39:31 - 39:37
it as a quadratic, so I want
positive square term and then
• 39:37 - 39:45
the linear term and then the
constant term. So I need to add.
• 39:45 - 39:51
This to both sides of 0 equals 2
sine squared X. Adding it to
• 39:51 - 39:58
both sides. Now I need to take
this away minus sign X from both
• 39:58 - 40:04
sides and I need to take the two
away from both sides. So one
• 40:04 - 40:06
takeaway two is minus one.
• 40:07 - 40:11
And now does this factorize?
• 40:11 - 40:17
look to see if we can make it
factorize 2 sign X and sign X.
• 40:17 - 40:21
Because multiplied together,
these two will give Me 2 sine
• 40:21 - 40:24
squared one and one because
multiplied together, these two
• 40:24 - 40:30
will give me one, but one of
them needs to be minus. To make
• 40:30 - 40:35
this a minus sign here. So I
think I'll have minus there and
• 40:35 - 40:39
plus there because two sign X
times by minus one gives me.
• 40:39 - 40:46
Minus 2 sign X one times by sign
X gives me sign X and if I
• 40:46 - 40:50
combine sign X with minus two
sign XI get minus sign X.
• 40:51 - 40:55
I have two numbers multiplied
together. This number 2 sign X
• 40:55 - 41:00
Plus One and this number sign X
minus one. They multiply
• 41:00 - 41:06
together to give me 0, so one or
both of them must be 0. Let's
• 41:06 - 41:07
write that down.
• 41:08 - 41:16
2 sign X Plus One is equal to
0 and sign X minus one is equal
• 41:16 - 41:23
to 0, so this tells me that sign
of X is equal. To take one away
• 41:23 - 41:30
from both sides and divide by
two. So sign X is minus 1/2 and
• 41:30 - 41:35
this one tells me that sign X is
equal to 1.
• 41:36 - 41:40
I'm now in a position to solve
these two separate equations.
• 41:41 - 41:43
So let me take this one first.
• 41:44 - 41:51
Now. We were working between
North and 2π, so we'll have a
• 41:51 - 41:53
sketch between North and 2π.
• 41:54 - 42:00
Of the sine curve and we want
sign X equals one. Well, there's
• 42:00 - 42:05
one and there's where it meets,
and that's pie by two, so we can
• 42:05 - 42:09
see that X is equal to pie by
• 42:09 - 42:16
two. Sign X equals minus 1/2.
Again, the range that we've been
• 42:16 - 42:22
given is between North and 2π.
So let's sketch between Norton
• 42:22 - 42:23
2π There's 2π.
• 42:25 - 42:27
Three π by 2.
• 42:28 - 42:34
Pie pie by two 0 - 1/2,
so that's coming along between
• 42:34 - 42:40
minus one and plus one that's
going to come along there.
• 42:41 - 42:46
And meet the curve there and
there. Now the one thing that we
• 42:46 - 42:49
do know is the angle who sign is
• 42:49 - 42:56
plus 1/2. Is π by 6, so we're
looking at plus 1/2. It will be
• 42:56 - 42:59
there and it would be pie by 6.
• 43:00 - 43:07
So it's π by 6 in from there,
so symmetry tells us that this
• 43:07 - 43:14
must be pie by 6 in from there,
so we've got X is equal to π
• 43:14 - 43:22
+ π by 6, and symmetry tells us
it's pie by 6 in. From there, 2π
• 43:22 - 43:24
- Π by 6.
• 43:25 - 43:33
There are six sixths in pie, so
that's Seven π by 6. There is
• 43:33 - 43:39
1216, two Pi. We're taking one
of them away, so it will be
• 43:39 - 43:41
11 Pi over 6.
• 43:42 - 43:47
So we've shown there how to
solve some trig equations.
• 43:47 - 43:52
The important thing is the
sketch the graph. Find the
• 43:52 - 43:57
initial value and then
workout where the others are
• 43:57 - 44:01
from the graphs. Remember,
the graphs are all symmetric
• 44:01 - 44:06
and they're all periodic, so
they repeat themselves every
• 44:06 - 44:08
2π or every 360 degrees.
Title:
www.mathcentre.ac.uk/.../5.5Solving%20Trigonometric%20Equations.mp4
Video Language:
English