Q: 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?

The two rules are equivalent. lim(𝑥 → 0) [(1 − cos 𝑥)∕𝑥] = 0 ⇔ ⇔ lim(𝑥 → 0) [−(cos 𝑥 − 1)∕𝑥] = 0 ⇔ ⇔ −lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0 ⇔ ⇔ lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0

Q: 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...

You could do an algebraic proof too. Sal already did one for sin(x) so he probably thought of spicing things up and hence did a more visual proof for cos(x)

Question 1

What is a wedge? 1st video, 2:56-2:58

Accepted Answer

A wedge is something that has one thick or large side decreasing to a thin edge, like a doorstop. In this case, the fraction of the unit circle which he highlighted goes from a large arc on the edge of the unit circle to a single point in the center.

Question 2

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?

Accepted Answer

(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 :-)

Question 3

in video 1 , why do we need 4 quadrant and absolute value? how does absolute value work?

Accepted Answer

The absolute value means the length must be positive because its in the quadrant I, (points and lines in quadrant I must be positive)

Question 4

in the third video at 4:00 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?

Accepted Answer

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!

Question 5

video one 3:00 , why is the area = theta/2pi x pi ? how and why this works , thanks !

Accepted Answer

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 3:00). 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.

Question 6

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).

Accepted Answer

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)

Question 7

At 4:11, 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?

Accepted Answer

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.

Question 8

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?

Accepted Answer

The two rules are equivalent.

lim(𝑥 → 0) [(1 − cos 𝑥)∕𝑥] = 0 ⇔
⇔ lim(𝑥 → 0) [−(cos 𝑥 − 1)∕𝑥] = 0 ⇔
⇔ −lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0 ⇔
⇔ lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0

Question 9

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...

Accepted Answer

You could do an algebraic proof too. Sal already did one for sin(x) so he probably thought of spicing things up and hence did a more visual proof for cos(x)

Course: AP®︎/College Calculus AB > Unit 2

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

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

1. $lim_{x \to 0} \frac{\sin (x)}{x} = 1$ ‍

2. $lim_{x \to 0} \frac{1 - \cos (x)}{x} = 0$ ‍

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

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

Want to join the conversation?

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 2

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

1. limx→0sin⁡(x)x=1‍

2. limx→01−cos⁡(x)x=0‍

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

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)‍ .

Want to join the conversation?

1. $lim_{x \to 0} \frac{\sin (x)}{x} = 1$ ‍

2. $lim_{x \to 0} \frac{1 - \cos (x)}{x} = 0$ ‍

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

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