I am learning about the derivative function of ddx[sin(x)]=cos(x).
The proof stated: From limx→0sin(x)x=1...
I realized I don't know why, so I wanted to learn why part is true first before moving on. But unfortunately I don't have the complete note for this proof.
- It started with a unit circle, and then drew a triangle at (1,tan(θ))
- It show the area of the big triangle is tanθ2
- It show the area is greater than the sector, which is θ2
Here is my question, how does this "section" of the circle equal to θ2? (It looks like a pizza slice). From there, it stated the area of the smaller triangle is sin(θ)2. I understand this part. Since the area of the triangle is 12(base×height).
Then they multiply each expression by 2sin(θ) to get
1cos(θ)≥θsin(θ)≥1
And the incomplete notes ended here, I am not sure how the teacher go the conclusion limx→0sin(x)x=1. I thought it might be something to do with reversing the inequality... Is the answer obvious from this point? And how does step #3 calculation works?
Answer
Draw the circle of radius 1 centered at (0,0) in the Cartesian plane.
Let θ be the length of the arc from (1,0) to a point on the circle. The radian measure of the corresponding angle is θ and the height of the endpoint of the arc above the coordinate axis is sinθ.
Now look at what happens when θ is infinitesimally small. The length of the arc is θ and the height is also θ, since that infinitely small part of the circle looks like a vertical line (you're looking at the neighborhood of (1,0) under a microscope).
Since θ and sinθ are the same when θ is infinitesimally small, it follows that sinθθ=1 when θ is infinitesimally small.
That is how Leonhard Euler viewed the matter in the 18th century.
Why does the sector of the circle have area θ/2?
The whole circle has area πr2=π12=π. The fraction of the circle in the sector is
arccircumference=θ2π.
So the area is
θ2π⋅π=θ2.
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