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## ← Derivatives of Sine and Cosine | MIT 18.01SC Single Variable Calculus, Fall 2010

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Derivatives of Sine and Cosine

Instructor: Joel Lewis

View the complete course: http://ocw.mit.edu/18-01SCF10

More courses at http://ocw.mit.edu

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Afficher la révision 1 créée 03/10/2011 par audreycermak.

1. Hi! Welcome back to recitation!
2. In the last lecture, we talked about finding the derivatives of trigonometric functions,
3. in particular the sine function and the cosine function.
4. So today, let's do an example of putting that into practice.
5. So, here's a function: h of x equal to sine of x, plus square root of three times cosine of x.
6. And I'm asking you to find which values of x have the property that the
7. derivative of h of x is equal to zero.
8. So, why don't you take a minute to think about that. Work that out on your own.
9. Pause the video and we'll come back and we'll work it out together.
10. All right, so, you've hopefully had a chance to look over this problem, try it out for yourself,
11. now let's see how to go about it.
12. So we have the function h of x. It's equal to sine x plus square root of three cosine x.
13. And we want to know when its derivative is equal to zero.
14. So in order to answer that question we should figure out what its derivative actually is.
15. And try and write down a formula for its derivative.
16. So in this case, that's not that bad. If we take a derivative of h,
17. well, h is a sum of two functions: sine x and square root of three cosine x.
18. And we know that the derivative of a sum is just the sum of the derivatives.
19. So we have the h prime of x is equal to d over dx sine x, plus
20. d over dx of square root of three times cosine x.
21. Now we learned last time in lecture that the derivative of sine x is cosine of x.
22. And we learned that the derivative of cosine of x is minus sine of x.
23. So here we have a constant multiple, but by the constant multiple rule
24. that just gets pulled out. So this is equal to cosine x minus square root of three times sine x.
25. So this is h prime of x. And now we want to solve the equation h prime of x equals zero.
26. So we want to find those values of x, such that cosine x minus
27. square root of three sine x is equal to zero.
29. I think my preferred way is I would add the square root of three sine x to one side,
30. and then I want to get my x's together so I would divide by cosine x.
31. So that gives me -- so on the left side I'll be left with cosine x divided by cosine x --
32. so that's just one. And on the right side I'll have the square root of three times sine x over cosine x,
33. so that's just square root of three times tan x.
34. Or, and I can rewrite this as tan of x is equal to one divided by
35. square root of three. Now to find x here, either you can remember your
36. special trig angles and know which values of x make this work,
37. or you could apply the arc tangent function here.
38. So in either case, the simplest solution here is x equals pi over six.
39. So if you like, you can draw a little right triangle. You know if this is x
40. -- if tan x is one over square root of three -- we should have this side being one
41. and this side being square root of three and in that case, in this right triangle
42. the hypotenuse would be two. And so then you would recognize this is a
43. 30 degree angle, or a pi over six radient angle.
44. But one thing to remember is that tangent of x is a periodic function
45. with period pi, so not only is pi over six a solution, but pi over six plus pi
46. is a solution. So that's seven pi over six. Or pi over six plus two pi.
47. Which is thirteen pi over six, or pi over six minus pi, which is minus five pi over six.
48. Etcetera, so there are actually infinitely many solutions --
49. they're given by pi over six plus an arbitrary multiple of pi.
50. So then, we're done! I do want to mention though that there's another
51. approach to this question. Which is we can start by multiplying
52. this expression for h of x. So if you look at h of x -- it's sine of x
53. plus square root of three cosine of x -- it resembles closely
55. So, in particular, to make it resemble it even more I can multiply
56. and divide by two. So I can rewrite h of x equals two times one half
57. sine x plus square root of three over two times cosine x.
58. And now one half is equal to cosine of pi over three and
59. square root of three over two is equal to sine of pi over three.
60. So I can rewrite this as two times cosine pi over three, sine x plus
61. sine pi over three, cosine x. And this is exactly what you get
62. when you do the angle addition formula for sine.
63. This is the expanded out form. And so we can apply it in reverse.
64. And get that this is equal to two times sine of x plus pi over three.
65. So far we haven't done any calculus. We've just done --
66. so in this solution -- our first solution -- we did some calculus first
67. and then some algebra and trigonometry. So, so far we've just done
68. some algebra and trigonometry. Now the points where h prime of x is equal to zero
69. are the points where the graph of this function has a horizontal tangent line.
70. So either you can compute its derivatiave using your rules, or by the definition.
71. Or you can just say, "Oh, we know what this graph looks like."
72. So I've sort of drawn a schematic up here. So this is a graph
73. -- this is the graph y equals two sine of x plus pi over three.
74. It's what you get if you take the graph y equals sine x and you shift it left by pi over three.
75. And you scale it up by a factor of two. So this, this here is f of x
76. equals minus pi over three. This root is x equals two pi over three.
77. And the points we are interested in are the points where there's a
78. horizontal tangent line - where the derivative is zero.
79. And so there's one of these right at this value, which is pi over six.
80. And the second one is this, is that minimum there.
81. So that happens at x equals seven pi over six.
82. Pi over six because for the usual sine function it happens at pi over two,
83. but we shifted everything left by pi over three. And so pi over two
84. minus pi over three is pi over six. And here for just y equals sine x,
85. this minimum would happen at three pi over two, but we've shifted
86. it left by pi over three. And so on. There's another trough over here,
87. and another peak over there and so on.
88. So that's the second way you can do this question using this cute trig identity here.
89. And that's that!