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
Proof: sin(a+b) = (cos a)(sin b) + (sin a)(cos b) Proof of the trig identity sin(a+b) = (cos a)(sin b) + (sin a)(cos b)
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- Welcome back.
- I'm now going to do a proof of a trig identity, which
- I think is pretty amazing.
- Although, I think, the proof isn't that obvious.
- And I'll have to admit ahead of time, this isn't something that
- would have occurred to me naturally.
- I wouldn't have naturally drawn this figure just
- to start off with.
- Let's just say we want to figure out some other way to
- write the sine of alpha plus beta, where alpha and beta are
- let's say, two separate angles.
- So if I had the sine of 40 and 50 degrees, I'd want to know--
- this would obviously be the sine of 90, which is easy.
- But could I rewrite that as some combination of the sine
- of 40 and the sine of 50 or whatever?
- I think you'll see where this is going.
- So let's go back to this diagram and let's say
- that this-- let me pick a better color.
- Let's say that this is angle alpha and this is angle beta.
- Than this whole angle right here is angle alpha plus beta.
- So we want to figure out the sine of alpha plus beta.
- Well, the sine of alpha plus beta, the sine of this
- whole angle, opposite over hypotenuse.
- Opposite this whole angle is if we use this right angle-- or
- this right triangle, triangle BAC.
- Opposite is BC, so that equals BC.
- I'll draw a little line over it.
- BC over the hypotenuse, AB.
- BC over AB is the sine of alpha plus beta.
- Well, can be write BC over AB differently?
- Let's see if we can.
- And probably, the person who first figured out this proof
- was just playing around.
- They drew this diagram, they said, can I write
- BC any differently?
- Well BC-- this whole length-- is the sum of BD and EF.
- And we know that because this is a horizontal line right now
- and you can figure that out just by looking at all
- the right angles.
- But this is a horizontal line.
- So BC is the same thing is BD plus EF.
- Let's write that one down.
- BC is the same thing as BD plus EF.
- And then still, all of that, over AB.
- All I did is I rewrote BC as a sum of this segment and this
- segment, which should make sense to you, hopefully.
- And then we can of course, rewrite that as equal to BD
- over AB plus EF over AB.
- So BD over AB plus EF over AB.
- And these are kind of nonsensical ratios, right?
- BD over AB, what can I do with that?
- And EF over AB, what can I do with that?
- Wouldn't it be more interesting if I could do like BD over BE.
- That'd be an interesting ratio because that would be a
- segment over its hypotenuse.
- So let's see if we can rewrite it somehow like that.
- Well, we could just do it mathematically.
- We could say this is equal to BD over BE times BE over AB.
- So this might seem non-intuitive to you, but
- it kind of makes sense.
- We didn't pick BE completely arbitrarily.
- We said we know what BD is, so let me pick another side that I
- can do something maybe with real trig ratios.
- And so I said BD over BE times BE over AB is
- equal to BD over AB.
- I hope I don't confuse you with all these letters.
- But that makes sense, right?
- Because these two terms would just cancel out.
- If we're just multiplying these fractions then you would
- get back to this top term.
- Let me actually make sure that you understand
- that this-- whoops.
- That this term and this term are the same thing.
- And now let's do that second term.
- We know EF, wouldn't it be good if we could relate EF to
- something, like it's the hypotenuse of this
- right triangle?
- Like AE.
- So let's do that.
- So let's put the plus sign there.
- EF over AB is the same thing as EF over AE times AE over AB.
- Once again, we're just multiplying fractions.
- These would cancel out and you would get this again.
- Let me make sure you understand that this term is the
- same thing as this term.
- And you can just multiple out the fractions and that's
- what you would get.
- Now before we progress with this whole line of
- thought that we're doing.
- Let's see if we could figure out something else interesting
- about this strange set of triangles and shapes
- that I've drawn.
- It's actually pretty neat.
- IF this angle is alpha-- we have line AF.
- EF is perpendicular to it, right?
- And DE is perpendicular to EF, right?
- So DE, this line, and AF are parallel.
- Since AF is parallel to DE and then, AE intersects both,
- we know that, what is that?
- The inner angles?
- Yeah, I think that's called inner angles
- with parallel lines.
- That this is also equal to alpha.
- You can imagine long parallel line here, long parallel here,
- and then this line intersects both.
- So if this is a little confusing maybe you want to
- review a little bit of the parallel line geometry, but I
- think this might make sense.
- So if this angle is alpha, then this angle right here
- is complementary to it.
- So it's 90 minus alpha.
- And if this angle is 90 minus alpha, this
- angle is obviously 90.
- Then we know that this angle plus this angle plus this
- angle has to equal 180.
- So we know that this is equal to alpha.
- If that doesn't make sense to you, think about this: alpha
- plus 90 minus alpha plus 90-- that's a minus.
- Minus alpha.
- Plus 90 is what?
- Alpha plus 90 minus alpha.
- So this minus alpha and alpha cancel out and you just have 90
- plus 90 and that equals 180.
- So we know that this angle right here, I know it's
- getting really small and probably hard to read.
- We know that this angle here is alpha.
- So let's get back to what we were progressing,
- what we were doing here.
- So what is BD over BE?
- BD over BE.
- Well, that's the adjacent to this alpha, which is
- the same angle really.
- BD over BE, so it's adjacent over hypotenuse.
- Cosine.
- So that is equal to the cosine of alpha.
- And what's BE over AB?
- Well, if we look at this larger right triangle, that is the
- opposite of beta times its hypotenuse.
- So what's opposite over hypotenuse?
- SOH.
- S O H.
- Sine.
- So sine of beta is BE over AB.
- So this is sine of beta.
- And now let me switch to magenta.
- What's EF over AE?
- If we look at this right triangle right here,
- is opposite over hypotenuse for alpha.
- So it's sine of alpha.
- Opposite over hypotenuse.
- And what's AE over AB?
- So now we're looking at this large right triangle here.
- AE over AB.
- Well, that's the adjacent of beta over the hypotenuse.
- Well, what's adjacent over hypotenuse?
- That's the cosine.
- CAH.
- Cosine of beta, of this beta right here.
- I think we're done.
- This is to me, fairly mind blowing.
- That the sine of alpha plus beta is equal to the cosine of
- alpha times the sine of beta.
- Plus the sine of alpha times the cosine of beta.
- What's neat about this is that it kind of came out of this
- nice symmetric formula.
- It's not this big, hairy thing.
- You might have even guessed it.
- I don't know.
- I just find it very neat.
- We went through this big convoluted proof with this big
- convoluted shape, but we got this nice symmetric trig
- identity out of it.
- So hopefully you found that amazing as well and in the next
- presentation I'll do a proof for cosine of alpha plus beta.
- See you soon.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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