Trigonometric identities
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Trigonometric Identities
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Pythagorean identities
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Proof: sin(a+b) = (cos a)(sin b) + (sin a)(cos b)
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Proof: cos(a+b) = (cos a)(cos b)-(sin a)(sin b)
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Trig identities part 2 (part 4 if you watch the proofs)
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Trig identies part 3 (part 5 if you watch the proofs)
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Addition and subtraction trig identities
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Law of cosines
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Law of cosines
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Navigation Word Problem
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Proof: Law of Sines
Navigation Word Problem Trigonometry problem involving two ships colliding.
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- I have a problem here from my cousin.
- Let's see what it says.
- It says Milwaukee, Wisconsin is directly west of Grand
- Haven, Michigan, on opposite sides of Lake Michigan.
- So let me draw that.
- So if we say that this right here is Milwaukee.
- It's due west of Grand Haven, Michigan.
- So if I were to draw-- let me draw a horizontal line.
- We're going over water so let me do it in blue.
- It's due west of Grand Haven, Michigan.
- You know, you go straight east.
- If I were to go straight east, I would get to
- Grand Haven, Michigan.
- Which is right here.
- Let me just label that, g.
- All right.
- On a foggy night, a law enforcement boat leaves
- from Milwaukee on a course of 105 degrees.
- The hardest part of these problems, in my opinion, is
- really just trying to figure out the convention
- they're using.
- When they say a course of 105 degrees, what does that mean?
- What direction is it?
- And I checked with my cousin, and her book says that in, I
- guess in the boating world, the course is how many
- degrees clockwise you're going, of due north.
- So due north is 0 degrees.
- So 105 degrees.
- So due north would be 0 degrees.
- Due north would be-- that's 0 degrees.
- So he's going 105 degrees clockwise of that.
- So if we 105 degrees-- so he's going 105 degrees.
- So that's like 105 degrees.
- Something like that.
- He's going 105 degrees.
- And I'll do him in magenta.
- His course is 105 degrees.
- So that's 105 degrees clockwise of due north.
- And what is that-- oh, it's like my screen just backed up.
- So what is that angle, in kind of what we're familiar with?
- Well, this would be 90 degrees right here.
- Going here.
- And then he goes 15 more degrees.
- So in kind of unit circle terms, this would be
- negative 15 degrees.
- Or if we wanted to figure out the angle of this vertex right
- here, that would be 90, and then we'd go another 15.
- So this angle right here is going to be what?
- It's going to be 15 degrees.
- Because he said his course is 105.
- So 90 plus 15 is 105.
- So how much south he's going of, kind of, straight
- west-east, is 15 degrees.
- This whole thing is 105.
- OK, let me keep reading the problem.
- He leaves from Milwaukee at a course of 105 degrees at the
- same time that a small smuggling craft steers a
- course of 195 degrees from Grand Haven.
- 195 degrees.
- So once again, due north is 0 degrees.
- And this guy's going 195.
- So we figure-- this is just a convention.
- You figure out-- well, he's going 195 degrees
- clockwise of due north.
- So 195 degrees is going to be-- let's see, it's going to be
- a 180 degrees and then some.
- It's going to be like that.
- So his course is going to look something like this.
- His course is going to look something like that.
- And let's see if we can figure out what this
- angle right here is.
- Because we, as you can kind of see where this is going, we're
- trying to figure out probably the size of this triangle,
- if I had to guess.
- I haven't even read the whole problem yet.
- Let's see.
- So he's going 195 degrees.
- So if we were to drop, well, like here.
- This right here is a 180 degrees, to go clockwise from
- straight up to straight down.
- This is 180.
- And so he went 195.
- So this is going to be-- that's going to be 15 degrees.
- And if this angle is 15 degrees, what is this
- angle going to be?
- Well, this entire angle is 90 degrees, right?
- It's kind of the third quadrant when we're thinking
- in unit circle terms.
- So this angle right here is going to be 90
- minus this 15 degrees.
- So what's 90 minus 15?
- It's 75, right?
- 75 degrees.
- And if we wanted to convert his, kind of, course angles
- into unit circles-- you know, with unit circles, you start
- here and you go all the way around this way.
- So I think you would get something like 255 degrees.
- But anyway.
- So we figured out that this angle is 15 degrees, this
- angle is 75 degrees.
- What's this angle going to be?
- This angle is going to be-- these all have to add
- up to 180, right?
- So this is going to be 180 minus 15 minus 75.
- And what's that?
- That's 180 minus 90.
- So 180 minus 90 is 90 degrees!
- So this angle here is 90 degrees.
- It's a right angle.
- It's a right angle.
- Interesting.
- OK, so what do they tell us?
- They tell us the law enforcement boat
- averages 23 knots.
- So he's traveling in this direction at 23 knots.
- All right, that's a little bit faster than 23 miles per hour.
- And collides with the smuggling craft.
- What was the smuggling boat's average speed?
- So they both leave their respective sites at the
- same time, and they both collide, right?
- So the time they traveled is the same.
- Let's call that time, t.
- Right?
- I don't know.
- They both left at the same time and it took some
- time for them to collide.
- So let's say that the time between when they left and
- the time they collided is t.
- So how far did the patrol boat travel?
- Well, he traveled at a speed of 23 knots, and it took him time,
- t, to get to the collision.
- So the distance he traveled is 23t.
- Speed times time is equal to distance.
- So the length of this side is 23t.
- Similarly, this guy, we don't know his speed.
- Let's call it, I don't know.
- Let's call it x.
- His speed is x.
- But the distance he travels is x times t.
- x times t, right?
- So that's the length of this side.
- So let's see if we can figure out what x is.
- So what do we know?
- We know a lot about this.
- We know this is a right triangle, et cetera.
- We know this angle.
- So if we wanted to solve for xt and use this 23t
- information, let's see.
- We know-- look at this angle.
- If we use the 75 degrees, we know the opposite angle.
- The opposite side, sorry.
- Which is 23t.
- And we know the adjacent side, which is xt.
- So let me write SOHCAHTOA here.
- So what deals with opposite and adjacent?
- Well, that's tangent, right?
- TOA.
- So if we say the tan of 75 degrees is going to be equal to
- the opposite side-- 23t-- over the adjacent side-- that's this
- side, opposite over adjacent-- xt.
- Well, the t's cancel out, right?
- The t's cancel out.
- And let's see if we can solve for x.
- Multiply x times both sides.
- You get x tangent of 75 is equal to 23.
- And then divide both sides by the tan of 75 and you get x is
- equal to 23 divided by the tangent of 75 degrees.
- And so that's our answer.
- And if I had to-- well, actually, I have-- let's see.
- Tangent of 75 degrees.
- I don't have a calculator in front of me.
- You could calculate it.
- It's actually going to be a pretty high number.
- So you could try to fit, you know.
- If you have a calculator, just type in 75 degrees.
- Take the tangent of it, and perform this calculation.
- But we've essentially solved this problem.
- I'll see you in the next video.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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