BAM!!! Mr. Tarrou. In this video we are going to take a look at solving trig equations again.
This time with multiple angles. This will be the second type of video I do like this.
But the first one, the difference is that it had a restriction on the values that I
was looking for. I wanted theta to be between 0 and 2pi. Here we are going to be looking
for all solutions. We are going to take a look at -2sin(x/2)... This is the multiple
angle that I am talking about where it is not just a sin(x), cos(x), tan(x) and so on...
It is (1/2)x. We have an adjustment on the angle size.... now minus the square root of
3. We only have one term with the trig function in it. We are going to isolate it and see
what we get. We are going to add the square root of 3 to both sides. We are going to divide
both sides by -2. We are going to get the sine function away from our angle, remember
it is an angle measure that we put into a trig function, by doing inverse sine. We get
x/2 equals the inverse sine of negative square root 3 over 2. If we were looking for answers
that were restricted between 0 and 2pi, this equation would not have a solution because
when you have a half angle in your trig function and you refer to your unit circle you are
only going to go half way around it. That way when you multiply by 2 you will have all
of your x angle measurements be between 0 and 2pi or 360 if you are using degrees. So
if we had a restriction on our values that we are allowed to give as answers this would
actually have no solution. But since we don't and we are looking for all possible solutions
we are going to go around this unit circle until we find those angle measures, that is
what you get from an inverse trig function, those angle measures that give you a sine
value of negative square root of 3 over 2. The sine function is y/r. So the y values
are on the unit circle. We have here at 4pi/3 and 5pi/3. Our angles are in quadrant 3 and
4 where our sine function is going to give you a negative ratio. So X/2 is equal to 4pi/3
and 5pi/3. We were not asked to solve for the angle measure that is inside the sine
function of x/2. We are trying to solve for x. So we are going to multiply everything
by 2 to get rid of this division of 2. Just put it all the way through... I guess you
could argue I have two equations here of x/2=4pi/3 and x/2=5pi/3. But anyway. We are going to
multiply everything by 2. We get x is equal to... 4 times 2 is 8pi/3... and 2 times 5ip/3
is 10pi/3. Leaving some space because I know that I am going to need to add something to
these answers to express that I am giving all possible answers, not just these two specific
angle measures. Please note again that these are outside the standard restricted values
of your answers for these questions of 0 to 2pi or 0 to 360. So what do I need to add
to these expressions to show that I am accepting all possible solutions? I need to do a plus....
Where is another color here?... I need to add in the period of the function times n.
Well what is the period? When you remember graphing like y=sin(x/2). If you were graphing
this you would be graphing y equals the sine of 1/2 x. That is probably what this would
look like in the section where you were graphing. This is the value of b. Period for sine, cosine,
secant, and cosecant is pi/b. For tangent and cotangent it is only pi/b. So the period
of this trig function of sin(x/2) or the sine of 1/2 x is 2pi divided by 1/2. The period...
Let me just write this. I was going to verbally say it. Let me write it out. 2pi divided by
1/2 is equal to, taking the 1/2 in the denominator and flipping it up to multiplication. So it
is 2pi multiplied the reciprocal of the denominator, the 1/2 flipped up, so the period is actually
going to be 4pi. So we are going to starting listing all of our possible solutions by taking
8pi/3 and saying it will be added by 4pi, the period of this trig function, time n.
n is an integer. n can be like -2, -1, 0, 1, 2, 3, 4, 5... It allows any full rotation
of this period. We are going to get more and more and more, an infinite number of angle
measures that will satisfy this equation. And the same thing with 10pi/3, add that by
the period of the function and again multiply it by n where n is an integer. So that is
example one, and I have one more coming. For our last example we have the sin(4x) plus
the sin(2x) equals 0. Now it looks like we should be able to solve this. We have got
the same trig function. It should be a little bit of factoring and we will be done. But
we have an angle of 4x and we have an angle of 2x. That might as well be a sine and a
cosine function. This normally... This is basically like having two variables in the
same equation. You cannot solve for two variables when there is only one equation, unless you
happen to be really really lucky and those two variables factor apart. We are going to
take a double angle identity, and that will allow us to reduce this angle measure of 4x
and bring it down to a measure of 2x. We are going to use the identity of sin(2x)=2cos(x)sin(x).
But of course this is not a 2x, it is 4x. That does not really matter that this is 1x
and this is 2x, what matters is this angle measure is double what we have on the other
side of the identity. This is the identity I am going to use except there is going to
be a 4x here, so these will be 2x. So this identity substitution, will get all of our
angle measures to be the same size. We are getting closer to getting ready to solve this.
But now we have got two different trig functions of sine and cosine. So still we have an issue
here with again the two variables in the same equation. That could be a problem except both
of these terms have a sin(2x). So we can factor that out, undistributed that sin(2x) and get
the sin(2x) times... the 2sin(2x)cos(2x)... take out the sin(2x) and we get 2cos(2x).
The sin(2x) divided by sin(2x) of course is 1. I have it factored. Now set each one equal
to zero. Get each trig function alone. Sine is already alone. So we are going to subtract
by 1 and divide by 2. I am going to get cos(2x) is equal to -1/2. We are going to get those
trig functions away from the angle measures by using the inverse trig functions. So 2x
is equal to the inverse sine of 0, and 2x is equal to the inverse cosine of -1/2. Now
if we were only concerned about getting value of x between 0 and 2pi or 0 and 360 if you
are doing degrees, then I would need to go around the unit circle twice and collect all
of the possible angles so that I could divide them by 2 and get all possible solutions between
0 and 2pi. But we are looking for all solutions and not just those within a restricted range.
So we are going to not worry about going around this unit circle twice. I am just going to
do this once, find the angles I need, and then add some little expressions at the end
to show that I want all possible solutions. Inverse sine of zero, where on this unit circle
is the sine value equal to 0? Where is the y equal to zero? 0 and pi. Over here we have
got... Excuse me. I knew I forgot something. 2x equals 0 and pi. Now we have 2x equals
the inverse cosine of -1/2. Where is my cosine equal to -1/2? Where are x values of -1/2?
At 2pi/3 and 4pi/3. We are solving for x so we are dividing everything by 2. So dividing
both sides by 2 we have x is equal to 0 and pi/2. Dividing everything by 2 over here,
or because we have fractions multiplying everything by 1/2. We get 2 times 1/2 is just simply
x. Then 1/2 times 2pi/3. The 2's are going to cancel out giving me pi/3. And I am running
out of room here. 1/2 times 4pi/3. 4 divided by 2 is equal to 2 so we have x=2pi/3. A little
sloppy there. We have the four solutions that we need. But I don't want four solutions,
I want all possible solutions. Let's think back to the previous example. We have a b
value of 2. We have a sine function and we have a cosine function. The period of both
of those trig functions is 2pi. So the period of 2pi/b. So each of these functions because
they are a double angle, if we were to graph them... they would oscillate and to a full
period within a horizontal distance of pi. Unlike the previous example we only have a
period of pi, so to express that you want all possible solutions you are going to go
to each one of these and add in that it is going to be this answer plus the period of
pi times n where n is an integer. When you are done with that, you will have listed all
of your possible solutions. So 0+pi(n), pi/2+pi(n), pi/3+pi(n), and 2pi/3+pi(n). There we are
done. We have answered or given all possible solutions to our last example. I am Mr. Tarrou.
BAM!!! Go Do Your Homework!