Hi! Welcome back to recitation!
In the last lecture, we talked about finding the derivatives of trigonometric functions,
in particular the sine function and the cosine function.
So today, let's do an example of putting that into practice.
So, here's a function: h of x equal to sine of x, plus square root of three times cosine of x.
And I'm asking you to find which values of x have the property that the
derivative of h of x is equal to zero.
So, why don't you take a minute to think about that. Work that out on your own.
Pause the video and we'll come back and we'll work it out together.
All right, so, you've hopefully had a chance to look over this problem, try it out for yourself,
now let's see how to go about it.
So we have the function h of x. It's equal to sine x plus square root of three cosine x.
And we want to know when its derivative is equal to zero.
So in order to answer that question we should figure out what its derivative actually is.
And try and write down a formula for its derivative.
So in this case, that's not that bad. If we take a derivative of h,
well, h is a sum of two functions: sine x and square root of three cosine x.
And we know that the derivative of a sum is just the sum of the derivatives.
So we have the h prime of x is equal to d over dx sine x, plus
d over dx of square root of three times cosine x.
Now we learned last time in lecture that the derivative of sine x is cosine of x.
And we learned that the derivative of cosine of x is minus sine of x.
So here we have a constant multiple, but by the constant multiple rule
that just gets pulled out. So this is equal to cosine x minus square root of three times sine x.
So this is h prime of x. And now we want to solve the equation h prime of x equals zero.
So we want to find those values of x, such that cosine x minus
square root of three sine x is equal to zero.
Now there are a couple different ways to go about this.
I think my preferred way is I would add the square root of three sine x to one side,
and then I want to get my x's together so I would divide by cosine x.
So that gives me -- so on the left side I'll be left with cosine x divided by cosine x --
so that's just one. And on the right side I'll have the square root of three times sine x over cosine x,
so that's just square root of three times tan x.
Or, and I can rewrite this as tan of x is equal to one divided by
square root of three. Now to find x here, either you can remember your
special trig angles and know which values of x make this work,
or you could apply the arc tangent function here.
So in either case, the simplest solution here is x equals pi over six.
So if you like, you can draw a little right triangle. You know if this is x
-- if tan x is one over square root of three -- we should have this side being one
and this side being square root of three and in that case, in this right triangle
the hypotenuse would be two. And so then you would recognize this is a
30 degree angle, or a pi over six radient angle.
But one thing to remember is that tangent of x is a periodic function
with period pi, so not only is pi over six a solution, but pi over six plus pi
is a solution. So that's seven pi over six. Or pi over six plus two pi.
Which is thirteen pi over six, or pi over six minus pi, which is minus five pi over six.
Etcetera, so there are actually infinitely many solutions --
they're given by pi over six plus an arbitrary multiple of pi.
So then, we're done! I do want to mention though that there's another
approach to this question. Which is we can start by multiplying
this expression for h of x. So if you look at h of x -- it's sine of x
plus square root of three cosine of x -- it resembles closely
one of your trigonometric identities that you know about.
So, in particular, to make it resemble it even more I can multiply
and divide by two. So I can rewrite h of x equals two times one half
sine x plus square root of three over two times cosine x.
And now one half is equal to cosine of pi over three and
square root of three over two is equal to sine of pi over three.
So I can rewrite this as two times cosine pi over three, sine x plus
sine pi over three, cosine x. And this is exactly what you get
when you do the angle addition formula for sine.
This is the expanded out form. And so we can apply it in reverse.
And get that this is equal to two times sine of x plus pi over three.
So far we haven't done any calculus. We've just done --
so in this solution -- our first solution -- we did some calculus first
and then some algebra and trigonometry. So, so far we've just done
some algebra and trigonometry. Now the points where h prime of x is equal to zero
are the points where the graph of this function has a horizontal tangent line.
So either you can compute its derivatiave using your rules, or by the definition.
Or you can just say, "Oh, we know what this graph looks like."
So I've sort of drawn a schematic up here. So this is a graph
-- this is the graph y equals two sine of x plus pi over three.
It's what you get if you take the graph y equals sine x and you shift it left by pi over three.
And you scale it up by a factor of two. So this, this here is f of x
equals minus pi over three. This root is x equals two pi over three.
And the points we are interested in are the points where there's a
horizontal tangent line - where the derivative is zero.
And so there's one of these right at this value, which is pi over six.
And the second one is this, is that minimum there.
So that happens at x equals seven pi over six.
Pi over six because for the usual sine function it happens at pi over two,
but we shifted everything left by pi over three. And so pi over two
minus pi over three is pi over six. And here for just y equals sine x,
this minimum would happen at three pi over two, but we've shifted
it left by pi over three. And so on. There's another trough over here,
and another peak over there and so on.
So that's the second way you can do this question using this cute trig identity here.
And that's that!
嗨！欢迎回来
在上一讲，我们谈到求解三角函数的导数
特别是正弦函数和余弦函数。
那么，今天，让我们做一个实际的例子。
因此，这里有一个函数：h(x)=sinx+√3 cosx
我要你们找出x的值是多少
当h(x)的导数等于0的时候
那么，你们自己花点时间来做一下
暂停视频，然后回来，我们一起做一下
好吧，希望你们已经自己思考并试着解了这个问题
现在让我们看看如何做。
我们有函数h(x)，它等于sinx+√3 cosx
我们想知道当它在什么情况下导数等于1
那么，为了回答这个问题，我们应该求出它的导数是多少
尝试并写下它的导数公式。
那么，在这里，这并难。如果想我们求h的导数
h是两个函数的总和：sinx和√3 cosx
而且我们知道和的导数就等于导数的和
那么h'(x)=d(sinx)/dx+d(√3 cosx)/dx
h'(x)=d(sinx)/dx+d(√3 cosx)/dx
那么，我们上次的课中讲到，sinx的导数就等于cosx
并且讲到，cosx 的导数就等于-sinx
在这里我们有一个常数乘，根据常数乘法则
只要把它提出来就可以了。那么这里等于cosx-√3 sinx
那么，这就是h'，现在我们来解h’(x)=0的等式
我们想解出x的值，使cosx-√3 sinx=0
cosx-√3 sinx=0
现在有几个不同的方法来解这个问题。
我认为我喜欢的方法是，把在一边加√3 sinx
然后我想找把x移到一遍，那么我会除以cosx
那么等到--在左边，剩下cosx，然后除以cosx
刚好等于1。而在右边得到√3 sinx/cosx
刚好等于√3tanx
或者，我可以把这个重写为tanx=1/√3
求这里的x，要么，你能记得
特殊三角函数值，知道这里x该等于什么值
或者你可以在这里应用反正切函数arc tan。
因此，在两种情况下，最简单的解是x=π/6
所以如果你喜欢，你可以画一个小直角三角形。
- 如果如果x=π/6话 - 这边就等于1
这边就等于√3，在这直角三角中
这斜边就等于2，你会认识到这是一个
30度角，或π/6弧度。
但有一点要记住的是，tan x是一个周期函数
以π为周期，所以这不是唯一的答案，
但π/6是其中一个解，还有7π/6，π/6+2π
30π/6，π/6-π，-5π/6等
那么，这里实际上有无穷多解 -
π/6 + π的任意倍数。
那么，我们就大功告成了！我也想提一提，
虽然有另一个方法解这个问题。既我们开始可以
h(x)乘以这个表达式，如果你看h(x)--等于
h(x)=sinx+√3 cosx
它接近你知道的三角恒等式
那么，我把它写的更像一些
我可以乘以2再除以2，我可以把h(x)改写为
1/2sinx+√3 cosx/2cosx
那么1/2等于cosπ/3
√3 /2=sinπ/3
因此，我可以把这个重写为2(cosπ/3sinx+sinπ/3cosx)
h(x)=2(cosπ/3sinx+sinπ/3cosx),这完全是
当你做sin时的加法公式
这是扩展公式，那么现在我们应用它
等到2sin(x+π/3)
到目前为止，我们还没有做任何微积分。我们刚刚做 -
所以在这个解决方案 - 我们的第一个解 - 我们先做了一些微积分
然后用一些代数和三角那么，到目前为止，我们只是做
一些代数和三角。现在，h(x)=0
意思是该函数的图形有一条水平切线。
那么你既可以应用法则或是定义来计算它的导数
或者你可以说：“哦，我们它的图形是什么样的。”
那么，我已经在这里画个图。
- 这是y=2（sinx+π/3)的图形
它是y=sin x 图形往左移动π/3
然后你把它扩大2倍。那么，这点等于-π/3
这点等于2π/3
而我们感兴趣的点是
有一个水平切线的点 - 在那里导数为零
那么这里有一个这样的值，等于π/6
第二个是这样的值在这里，它也是这里最小的值
那么，这种情况发生在点x=7π/6
π/6，因为，对函数sinx来说发生在π/2点
但我们我这整个往右移动了π/3.所以π/2-π/3=π/6
π/2-π/3=π/6，在这对对于y=sinx来说
最小值发生在3π/2，但我们这里
我们把它往左π/3，等等，当然这里还有另外一个值
和另一个峰值等等
那么这是第二种方法，用这个可爱的三角恒等式来解
今天就到这里！