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PROFESSOR: Hi.

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Welcome back to recitation.

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In last lecture we talked about
finding the derivatives

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of trigonometric functions.

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In particular, the
sine function

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and the cosine function.

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So today let's do an example of
putting that into practice.

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So here's a function h of x
equal to sine of x plus square

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root of 3 times cosine of x.

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And I'm asking you to find which
values of x have the

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property that the derivative
of h of x is equal to 0.

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So why don't you take a minute
to think about that, work it

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out on your own, pause the
video, and we'll come back and

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we'll work it out together.

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All right.

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So you've hopefully had a
chance to look over this

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problem, try it out
for yourself.

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Now let's see how
to go about it.

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So we have the function
h of x--

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it's equal to sine x plus square
root of 3 cosine x--

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and we want to know when its
derivative is equal to 0.

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So in order to answer that
question we should figure out

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what its derivative actually
is and try and write down a

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formula for its derivative.

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So in this case that's
not that bad.

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If we take a derivative of
h, well h is a sum of two

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functions--

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sine x and square root
of 3 cosine x.

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And we know that the derivative
of a sum is just

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the sum of the derivatives.

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So we have the h prime of x is
equal to d over dx of sine x

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plus d over dx of square root
of 3 times cosine x.

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Now we learned last time in
lecture that the derivative of

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sine x is cosine of x, and we
learned the derivative of

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cosine of x is minus sine x.

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So here we have a constant
multiple, but by the constant

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multiple rule that just
gets pulled out.

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So this is equal to cosine
x minus square root of

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3 times sine x.

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So this is h prime of x and
now we want to solve the

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equation h prime
of x equals 0.

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So we want to find those values
of x such that cosine x

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minus square root of 3
sine x is equal to 0.

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Now there are a couple
different

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ways to go about this.

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I think my preferred way is I
would add the square root of 3

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sine x to one side, and then I
want to get my x's together,

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so I would divide by cosine x.

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So that gives me, so on the left
side I'll be left with

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cosine x divided by cosine
x, so that's just 1.

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And on the right side I'll have
square root of 3 times

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sine x over cosine x.

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So that's just square
root of 3 times 10x.

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Or, and I can rewrite this as 10
of x is equal to 1 divided

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by square root of 3.

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Now to find x here, either you
can remember your special trig

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angles and know which values
of x make this work.

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Or you could apply the arc
tangent function here.

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So in either case, the simplest
solution here is x

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equals pi over 6.

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So if you like, you can draw
a little right triangle.

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You know, if this is x, if tan
x is 1 over square root of 3,

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we should have this side being
1 and this side being

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square root of 3.

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And in OK, in that case, then
this right triangle the

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hypotenuse would be 2.

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And so then you, you know, would
recognize this is a 30

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degree angle, or pi over
6 radian angle.

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But one thing to remember is
that tangent of x is a

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periodic function with period
pi, so not only is pi over 6 a

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solution, but pi over 6
plus pi is a solution.

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So that's 7 pi over 6, or pi
over 6 plus 2 pi, which is 13

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pi over 6, or pi over 6 minus
pi, which is minus 5 pi over

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6, et cetera.

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So there are actually infinitely
many solutions.

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They're given by pi
over 6 plus an

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arbitrary multiple of pi.

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So then we're done.

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So I do want to mention, though,
that there's another

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approach to this question,
which is, we can start by

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multiplying this expression
for h of x.

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So if you look at h of x--
that's sine of x plus square

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root of 3 times cosine x--

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it resembles closely one of your
trigonometric identities

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that you know about.

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So in particular, to make it
resemble it even more I can

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multiply and divide by 2.

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So I can rewrite h of x equals 2
times 1/2 sine x plus square

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root of 3 over 2
times cosine x.

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And now 1/2 is equal to
cosine of pi over 3.

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And square root of 3 over 2 is
equal to sine of pi over 3.

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So I can rewrite this
as 2 times--

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what did I say--

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I said cosine pi over 3 sine x
plus sine pi over 3 cosine x.

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And this is exactly what you
get when you do the angle

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addition formula for sine.

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This is the expanded out form,
and so we can apply it in

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reverse and get that this is
equal to 2 times sine of x

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plus pi over 3.

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So, so far we haven't
done any calculus.

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We've just done--

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so in this solution, our first
solution we did some calculus

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first and then some algebra
and trigonometry.

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So, so far we've just done some
algebra and trigonometry.

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Now the points where h prime
of x is equal to 0 are the

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points where the graph of this
function has a horizontal

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tangent line.

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So either you can compute its
derivative using your rules or

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by the definition.

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Or you can just say,
oh, we know what

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this graph looks like.

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So I've sort of drawn
a schematic up here.

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So this is a graph,
this graph is--

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OK, so this is the graph,
y equals 2 sine of x

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plus pi over 3.

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It's what you get if you take
the graph, y equals sine x,

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and you shift it left by pi over
3 and you scale it up by

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a factor of 2.

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So this here is at x equals
minus pi over 3.

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This root is x equals
2 pi over 3.

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And the points we're interested
in are the points

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where there's a horizontal
tangent line, where the

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derivative is 0.

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And so there's one of these
right at this value,

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which is pi over 6.

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And then the second one is this,
is that minimum there.

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So that happens at x
equals 7 pi over 6.

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pi over 6 because for the usual
sine function it happens

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at pi over 2, but we shifted
everything left by pi over 3.

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And so pi over 2 minus pi
over 3 is pi over 6.

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And here for this, for just y
equals sine x, this minimum

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would happen at 3 pi over
2, but we've shifted it

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left by pi over 3.

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And so on. you know, every,
there's another trough over

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here, and another peak over
there, and so on.

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So that's the second way you
can do this question using

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this cute trig identity here.

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And that's that.

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