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Proving the derivatives of sin(x) and cos(x)

​Proving that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x).
The trigonometric functions sin(x) and cos(x) play a significant role in calculus. These are their derivatives:
ddx[sin(x)]=cos(x)ddx[cos(x)]=sin(x)
The AP Calculus course doesn't require knowing the proofs of these derivatives, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

First, we would like to find two tricky limits that are used in our proof.

1. limx0sin(x)x=1

Khan Academy video wrapper
Limit of sin(x)/x as x approaches 0See video transcript

2. limx01cos(x)x=0

Khan Academy video wrapper
Limit of (1-cos(x))/x as x approaches 0See video transcript

Now we are ready to prove that the derivative of sin(x) is cos(x).

Khan Academy video wrapper
Proof of the derivative of sin(x)See video transcript

Finally, we can use the fact that the derivative of sin(x) is cos(x) to show that the derivative of cos(x) is sin(x).

Khan Academy video wrapper
Proof of the derivative of cos(x)See video transcript

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  • leafers sapling style avatar for user Kuzma L
    What is a wedge? 1st video, -
    (10 votes)
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  • blobby green style avatar for user mohamad
    in the third video at why did he say that"we can take cos(x)outside of limit"?cos(x) is not a constant.which property allow us to do that?
    (4 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      With respect to the quantity that is actually changing in the limit, namely delta x, cos(x) is a constant and so can be taken outside of the limit.

      If this is not clear, delta x could be called something else, say h, to make it more clear that cos(x) is considered a constant in this limit and so can be taken outside of the limit.

      Have a blessed, wonderful day!
      (31 votes)
  • blobby green style avatar for user Steph
    In the visual graph from the final video, since you are shifting left, wouldn't it be minus Pi/2, not plus? Visually, shouldn't it shift right? Albeit for the same effect. Would this mean that sine of x minus Pi over 2 is also cosine x?
    (6 votes)
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    • starky sapling style avatar for user tejas_gondalia
      (Edit): Because the original form of a sinusoidal equation is y = Asin(B(x - C)) + D , in which C represents the phase shift.
      So, here in this case, when our sine function is sin(x+Pi/2), comparing it with the original sinusoidal function, we get C=(-Pi/2). Hence we will be doing a phase shift in the left.
      So is the case with sin(x-Pi/2), in which we get C as Pi/2, hence the graph shifts towards the right.
      I'm sorry if I was not clear enough but this a seperate topic in itself which I can't explain in here.
      Hope this helps :-)
      (8 votes)
  • blobby green style avatar for user Abdulrahman Alothman
    video one , why is the area = theta/2pi x pi ? how and why this works , thanks !
    (3 votes)
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    • aqualine ultimate style avatar for user itimespi
      First off, we know the area a circle is pi * r². Since we are working with the unit circle, where the radius is 1, we could say that the area of this circle is simply pi.

      Now, at this moment, Sal is trying to find the area of a wedge of the circle, or a fraction of the entire circle's area, which is pi. So how do we calculate the area when we are given theta? Well, theta represents a radian measure, and that radian measure can tell us how much of the circle is being occupied by a wedge made with central angle theta (as shown in the picture at ). We also know that a full circle has a central angle of 2pi. When theta = 2pi, that means that the wedge occupies the entire circle. When theta = pi, that means that the wedge occupies half of the circle, and so on. Thus, the fraction of the circle being occupied by a wedge with central angle theta is theta/2pi, or the given radian measure divided by the radian measure when the wedge occupies the entire circle.

      Now, we have our fraction, and the area occupied by this fraction is found by multiplying the fraction and area. Thus, the area of our wedge is theta/2pi * pi = theta/2.
      (8 votes)
  • leafers tree style avatar for user areeb09871234
    in video 1 , why do we need 4 quadrant and absolute value? how does absolute value work?
    (4 votes)
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  • purple pi purple style avatar for user The first integral proponent
    Why derivative of sinx=limit of (sin(x+ Δx)-sinx)/x in the 3rd video? -
    (3 votes)
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  • blobby green style avatar for user jovancicluka
    Can anyone show an algebraic proof that the derivative of cos(x) = -sin(x)? When i used a similar method as seen in the proof of d/dx(sin(x))=cos(x) my result was that the derivative was a positive sin(x).
    (2 votes)
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    • leaf green style avatar for user kubleeka
      Recall that cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

      We're evaluating the limit as h goes to 0 of
      [cos(x+h)-cos(x)]/h
      [cos(x)cos(h)-sin(x)sin(h)-cos(x)]/h
      [cos(x)(cos(h)-1)-sin(x)sin(h)]/h
      cos(x)[(cos(h)-1)/h] -sin(x)[sin(h)/h]

      The two terms in square brackets are the special limits proven earlier in the playlist; the first is 0 and the second is 1, so the expression reduces to
      cos(x)·0-sin(x)·1
      -sin(x)
      (6 votes)
  • aqualine ultimate style avatar for user Eragon
    Is the rule 1-cos(x)/x = 0, the same as cos
    (x)-1/x=0? Some book are using cos(x) -1 while others are using 1-cos(x). Why?
    (1 vote)
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  • stelly green style avatar for user Talia
    Why didn't Sal just do an algebraic proof for finding the derivative of cos(x)? I just did it on paper myself in less than 5 minutes, using his outline for the proof of the derivative of sin(x) and the cosine sum formula. The visual proof is more intuitive, but the algebraic proof is solid. Interesting choice...
    (1 vote)
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  • blobby green style avatar for user Mateo Sabando
    At , I don't understand why Sal uses "less than or equal" rather than just "less than". When would the areas be equal? At theta=0?
    (2 votes)
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    • leaf green style avatar for user kubleeka
      Exactly, the areas are equal at θ=0. Also, the statement of the squeeze theorem involves non-strict inequalities (≤), not strict ones (<). So if we want to invoke the squeeze theorem, we need to use non-strict inequalities.
      (3 votes)