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i've already made a handful
of videos that covers what
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I'm going to cover, the
trigonometric identities I'm
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going to cover in this video.
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The reason why I'm doing it is
that I'm in need of review
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myself because I was doing some
calculus problems that required
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me to know this, and I have
better recording software now
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so I thought two birds with one
stone, let me rerecord a
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video and kind of refresh
things in my own mind.
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So the trig identities that I'm
going to assume that we know
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because I've already made
videos on them and they're a
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little bit involved to remember
or to prove, are that the sine
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of a plus b is equal to the
sine of a times the cosine
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of b plus the sine of b
times the cosine of a.
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That's the first one, I assume,
going into this video we know.
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And then if we wanted to know
the sine of-- well, I'll just
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write it a little differently.
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What if I wanted to figure out
the sine of a plus-- I'll
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write it this way-- minus c?
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Which is the same thing
as a minus c, right?
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Well, we could just use this
formula up here to say well,
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that's equal to the sine of a
times the cosine of minus c
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plus the sine of minus c
times the cosine of a.
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And we know, and I guess this
is another assumption that
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we're going to have to have
going into this video, that the
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cosine of minus c is equal
to just the cosine of c.
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That the cosine is
an even function.
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And you could look at that by
looking at the graph of the
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cosine function, or even at
the unit circle itself.
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And that the sine is
an odd function.
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That the sine of minus
c is actually equal
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to minus sine of c.
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So we can use both of that
information to rewrite the
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second line up here to say that
the sine of a minus c is equal
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to the sine of a times
the cosine of c.
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Because cosine of minus
c is the same thing
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as the cosine of c.
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Times the cosine of c.
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And then, minus the sine of c.
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Instead of writing this,
I could write this.
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Minus the sine of c
times the cosine of a.
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So that we kind of pseudo
proved this by knowing this
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and this ahead of time.
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Fair enough.
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And I'm going to use all of
these to kind of prove a bunch
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of more trig identities
that I'm going to need.
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So the other trig identity is
that the cosine of a plus b is
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equal to the cosine of a-- you
don't mix up the cosines and
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the sines in this situation.
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Cosine of a times
the sine of b.
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And this is minus--
well, sorry.
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I just said you don't mix it
up and then I mixed them up.
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Times the cosine of b minus
sine of a times the sine of b.
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Now, if you wanted to know what
the cosine of a minus b is,
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well, you use these
same properties.
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Cosine of minus b, that's still
going to be cosine on b.
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So that's going to be the
cosine of a times the cosine--
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cosine of minus b is the
same thing as cosine of b.
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But here you're going to have
sine of minus b, which is the
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same thing as the
minus sine of b.
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And that minus will cancel that
out, so it'll be plus sine
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of a times the sine of b.
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So it's a little tricky.
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When you have a plus sign
here you get a minus there.
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When you don't minus
sign there, you get
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a plus sign there.
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But fair enough.
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I don't want to dwell on that
too much because we have many
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more identities to show.
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So what if I wanted an
identity for let's
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say, the cosine of 2a?
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So the cosine of 2a.
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Well that's just the same thing
as the cosine of a plus a.
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And then we could use this
formula right up here.
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If my second a is just my b,
then this is just equal to
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cosine of a times the cosine
of a minus the sine of
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a times the sine of a.
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My b is also an a in this
situation, which I could
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rewrite as, this is equal to
the cosine squared of a.
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I just wrote cosine of a times
itself twice or times itself.
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Minus sine squared of a.
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This is one I guess
identity already.
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Cosine of 2a is equal to the
cosine squared of a minus
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the sine squared of a.
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Let me box off my identities
that we're showing
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in this video.
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So I just showed you that one.
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What if I'm not satisfied?
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What if I just want it
in terms of cosines?
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Well, we could break out
the unit circle definition
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of our trig functions.
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This is kind of the most
fundamental identity.
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The sine squared of a
plus the cosine squared
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of a is equal to 1.
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Or you could write that--
let me think of the
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best way to do this.
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You could write that the sine
squared of a is equal to
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1 minus the cosine
sign squared of a.
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And then we could take this
and substitute it right here.
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So we could rewrite this
identity as being equal to the
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cosine squared of a minus
the sine squared of a.
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But the sine squared of
a is this right there.
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So minus-- I'll do it
in a different color.
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Minus 1 minus cosine
squared of a.
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That's what I just substituted
for the sine squared of a.
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And so this is equal to the
cosine squared of a minus 1
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plus the cosine squared of a.
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Which is equal to--
we're just adding.
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I'll just continue
on the right.
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We have 1 cosine squared of a
plus another cosine squared
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of a, so it's 2 cosine
squared of a minus 1.
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And all of that is
equal to cosine of 2a.
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Now what if I wanted to get
an identity that gave me
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what cosine squared of
a is in terms of this?
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Well we could just
solve for that.
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If we add 1 to both sides of
this equation, actually,
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let me write this.
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This is one of our
other identities.
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But if we add 1 to both sides
of that equation we get 2 times
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the cosine squared of a is
equal to cosine of 2a plus 1.
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And if we divide both sides of
this by 2 we get the cosine
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squared of a is equal to 1/2--
now we could rearrange these
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just to do it-- times 1
plus the cosine of 2a.
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And we're done.
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And we have another identity.
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Cosine squared of a, sometimes
it's called the power reduction
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identity right there.
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Now what if we wanted
something in terms of
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the sine squared of a?
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Well then maybe we could go
back up here and we know from
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this identity that the sine
squared of a is equal to 1
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minus cosine squared of a.
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Or we could have
gone the other way.
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We could have subtracted sine
squared of a from both sides
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and we could have gotten--
let me go down there.
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If I subtracted sine squared of
a from both sides you could get
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cosine squared of a is equal
to 1 minus sine squared of a.
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And then we could go back into
this formula right up here and
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we could write down-- and I'll
do it in this blue color.
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We could write down that the
cosine of 2a is equal to--
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instead of writing a cosine
squared of a, I'll write this-
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is equal to 1 minus sine
squared of a minus
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sine squared of a.
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So my cosine of 2a is equal to?
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Let's see.
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I have a minus sine squared
of a minus another
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sine squared of a.
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So I have 1 minus 2
sine squared of a.
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So here's another identity.
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Another way to write
my cosine of 2a.
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We're discovering a lot of ways
to write our cosine of 2a.
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Now if we wanted to solve for
sine squared of 2a we could
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add it to both sides
of the equation.
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So let me do that and I'll
just write it here for
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the sake of saving space.
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Let me scroll down
a little bit.
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So I'm going to go here.
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If I just add 2 sine squared
of a to both sides of this, I
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get 2 sine squared of a plus
cosine of 2a is equal to 1.
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Subtract cosine of
2a from both sides.
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You get 2 sine squared of a is
equal to 1 minus cosine of 2a.
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Then you divide both sides of
this by 2 and you get sine
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squared of a is equal to 1/2
times 1 minus cosine of 2a.
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And we have our other discovery
I guess we could call it.
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Our finding.
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And it's interesting.
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It's always interesting
to look at the symmetry.
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Cosine squared-- they're
identical except for you have a
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plus 2a here for the cosine
squared and you have a minus
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cosine of 2a here for
the sine squared.
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So we've already found a
lot of interesting things.
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Let's see if we can do
anything with the sine of 2a.
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Let me pick a new color
here that I haven't used.
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Well, I've pretty much
used all my colors.
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So if I want to figure out the
sine of 2a, this is equal
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to the sine of a plus a.
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Which is equal to the sine of a
times the co-- well, I don't
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want to make it that thick.
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Times the cosine of a plus--
and this cosine of a,
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that's the second a.
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Actually, you could
view it that way.
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Plus the sine-- I'm just
using the sine of a plus b.
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Plus the sine of the
second a times the
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cosine of the first a.
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I just wrote the same thing
twice, so this is just people
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to 2 sine of a, cosine of a.
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That was a little bit easier.
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So sine of 2a is equal to that.
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So that's another result.
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I know I'm a little bit tired
by playing with all of
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these sine and cosines.
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And I was able to get all the
results that I needed for my
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calculus problem, so hopefully
this was a good review for
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you because it was a
good review for me.
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You can write these
things down.
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You can memorize them if you
want, but the really important
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take away is to realize that
you really can derive all of
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these formulas really from
these initial formulas
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that we just had.
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And even these, I also have
proofs to show you how to get
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these from just the basic
definitions of your
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trig functions.