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)

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.