00:01
For this problem, we're asked to do some work with phasers.
00:04
But first, we're just asked to use some trig identities.
00:08
So we've got y1 is 12 sign of 4 .5t.
00:21
And we've got y2 is 12 sign of 4 .5t plus 70 degrees.
00:35
And we're supposed to add these together first using a trig identity.
00:42
So we're adding two signs together.
00:49
So there is a trig identity in table b .3.
00:55
It's on page a12.
01:03
Okay.
01:04
And i guess i'll write down what it says.
01:06
It says the sign of a plus b equals the sign of a.
01:18
Cosine of b.
01:20
No, that's not the one i wanted.
01:23
Sorry about that.
01:29
Sign of a plus sign of b is two sign of one half a plus b.
01:50
Two sign of one half a plus b cosine of one half a minus b.
02:04
So our answer to this question then would be 12 centimeters times two.
02:27
So i might as well have just written 24 centimeters times the sign of one half of a plus b.
02:47
Okay.
02:48
Well, that's just the average of a plus b.
02:52
So that would be t plus 35 degrees.
03:06
Okay, good.
03:09
Cosine of 1 half a minus b.
03:19
Well, that would be negative 35 degrees.
03:30
Okay, so now i can just use the cosine of negative 35 degrees.
03:36
Make sure i'm in degree mode, which i am now.
03:40
2 sine of negative 35 degrees is .819.
03:50
I want to multiply that times 24.
03:59
19 .7 sign of 4 .5 t plus 35 degrees.
04:19
Now i'm going to add them in a phaser diagram.
04:23
I'm going to do it like this.
04:29
So we have one phaser that is 12 centimeters.
04:40
And we have another phaser.
04:42
Whoops, i wanted to do that a different color.
04:46
That is also 12 centimeters, but at 70 degrees.
04:55
So i can rewrite that green phaser over here.
05:04
I can show the resultant here in blue, the using the law of cosine.
05:12
C squared equals a squared plus b squared minus 2ab cosine of theta, which theta will be right in here, where theta is now 180 minus 70 degrees, and a and b are both 12 centimeters.
05:45
Putting that into a calculator, i get that c is 19 .7 centimeters.
05:57
Now, i also notice that since these sides are the same, that this side and this side are the same, so these angles here are the same.
06:13
And so by alternate interior angles, this angle is also the same.
06:19
And so that's an angle bisector.
06:22
And so the angle here, phi, has to be 35 degrees.
06:31
So our answer then becomes this, using the 19 .7 and the 35 degrees.
06:45
All right.
06:47
Now i need to add three waves.
06:51
So this is, i think, a totally different problem, but i notice that everything is the same.
07:01
We have one that is at, oh no, the magnitudes are the same, are different.
07:09
My bad.
07:10
So, let me see here.
07:20
I'm going to start by trying to do it this way, and we'll see if it works out.
07:33
We've got x and y here.
07:38
First, we've got 12 centimeters at an angle of 70 degrees.
07:47
Okay, so i'm going to call that...