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sin
(x) = cos(x)
Calculus 11, Veritas Prep.
We have long suspected, based on the visuals alone, that the derivative of sin(x) looks kind of like cos(x) (potentially with a vertical expansion/compression):
And I’ve assured you that, in fact, the derivative of sin(x) is cos(x). But how do we prove this? We tried to do this back in November^1 , but we ran into a stumbling block—one that caused us to trip and fall not onto the ground, but into a weird alternate universe of limits and epsilons and deltas and h’s going to zero. It gave us a whole new perspective—a fuller perspective—on what derivatives are. Now that we’re back in the main universe (the one that has derivatives), let’s see if we can actually prove this to be true. (That is, let’s see if we can see why this is true.) We want to calculate sin′(x), and since we have no other rules that we can apply (can’t use the product rule, can’t use the chain rule, etc.), we may as well start with first principles: plug it into Fermat’s difference quotient.
sin′(x) = lim h→ 0
[
sin(x + h) − sin(x) h
]
But this looks kind of nasty. Unlike with, e.g., x^2 , we can’t “multiply out” the sin(x + h), or distribute it. It’s certainly not equal to sin(x) + sin(h). But we can simplify it using that formula we proved in trig:
sin(α + β) = cos(α) sin(β) + sin(α) cos(β)
If we apply that here, we’ll get:
sin′(x) = lim h→ 0
[
sin(x + h) − sin(x) h
]
= lim h→ 0
[
cos(x) sin(h) + sin(x) cos(h) − sin(x) h
]
which, if I do some rearranging, is
= lim h→ 0
[
cos(x) sin(h) h
sin(x) cos(h) − sin(x) h
]
= lim h→ 0
[
cos(x)
sin(h) h
cos(h) − 1 h
]
= lim h→ 0
[
cos(x) sin(h) h − sin(x) − cos(h) + 1 h
]
(factoring out a − 1)
(^1) see the original limits notes
= lim h→ 0
[
cos(x)
sin(h) h − sin(x)
1 − cos(h) h
]
= lim h→ 0
[
cos(x)
sin(h) h
]
− lim h→ 0
[
sin(x)
1 − cos(h) h
]
= cos(x) · lim h→ 0
[
sin(h) h
]
− sin(x) · lim h→ 0
[
1 − cos(h) h
]
So now we’re basically at the place where we had to stop when we tried to do this in November. Because: what are these limits? It’s not clear how we work out limh→ 0 sin(h)/h and limh→ 0 (1 − cos h)/h, because if we simply plug 0 in, we get 0/0. Each of these functions has a hole at h = 0. So we need some way of finding the presumptive y-value of that hole—of finding what sin(0)/ 0 should be, if it existed. With rational functions, this is no big deal—we simply cancel out the offending/hole-creating factor, and then act as if it never existed. But we can’t do that here, because sine and cosine aren’t polynomials. So. Um. Geometry to the rescue! (I have this image of Mr. Austin swooping in with a cape.) We need to figure out what those two limits are. So we’ll go on a bit of a tangent, and build some triangles. What we’re going to do, in a sentence, is use geometry to build some algebraic relationships, and then use those to find our limit. We’re going to construct an inequality that has sin(x)/x in the middle (like stuff < sin(x)/x < more stuff), and then we’re going to (more or less) crush sin(x)/x between the two things on the outside. We’ll be able to find the limit as h → 0 of both of the things on the outside—they’ll both go to 1—and since we know that sin(x)/x is always between them, we’ll know that the limit as h → 0 of sin(x)/x will be 1, too. This is the general idea. So. Imagine I draw a right triangle with an angle of x, legs of length a and b, and hypotenuse of 1:
so then, just using trig, we know that
sin(x) = opp hyp
a 1
= a
and
cos(x) = adj hyp
b 1
= b
So rather than using two new variables, a and b, I can just label the legs with sin(x) and cos(x), since that’s how long each of them are:
Let’s call this the “big triangle”. I’ve labelled the side opposite x as γ, but we can write this solely in terms of x: tan(x) = opp adj
γ 1
= γ
So then rather than calling the far-right side γ, let’s call it tan(x):
Why are we doing all of this? We have three overlapping shapes: a small triangle, a pizza slice, and
a big triangle. They’re stacked on top of each other. We know that the little triangle is smaller than the pizza slice, and we know that the pizza slice is smaller than the big triangle. So we know:
area of little triangle < area of pizza slice < area of big triangle
But we also know what each of their areas are. The little triangle we know has a base of 1, and a height of sin(x) (from what we started with), so we can find its area. The big triangle has a base of 1, and a height of tan x. And the pizza slice—well, we know that it’s a slice out of a pizza with radius 1, and so the entire pizza will have area π · (1)^2. But it’s only a percentage of the entire pizza. It’s x radians of the entire pizza, and the entire pizza has 2π radians. So the pizza slice’s area will be (π · 12 ) · (x/ 2 π). In summary:
area of little triangle =
· base · height =
· 1 · sin x =
sin x
area of pizza slice = (area of full circle) · (percentage of the circle it is) = (π · 12 ) ·
( (^) x 2 π
x
area of big triangle =
· base · height =
· 1 · tan x =
tan x =
sin x cos x
So because we know:
area of little triangle < area of pizza slice < area of big triangle
if we plug in the areas, we must have:
1 2
sin(x) <
x <
sin x cos x
Let’s clean this up a bit. We want to get sin(x)/x in the middle there, so that we can CRUSH it. If I get rid of the 1/2s...
sin x < x < sin x cos x and divide by sin(x)... sin x sin x
x sin x
sin x cos x sin x
1 < x sin x
cos x
and then take the reciprocal (which will flip the direction of the inequalities^2 ):
1 1
sin x x
cos x 1
1 > sin x x
cos x
So I get a sin(x)/x! This is the thing whose limit as h → 0 we want to figure out. Here’s the trick: we know that sin(x)/x is always between 1 and cos x. And we know what happens to cos x and 1 as x → 0. Both of them just become 1. (Well, 1 already is 1):
cos(x) −−−x−→ 0
1 −−−x−→ 0
(^2) since (for example) 2 < 3 and^1 2
1 3
But we also know that we can split limits up along multiplication, and we know how to take the limit of both of those things! We just found out that as h → 0, sin(x)/x goes to 1, and as h → 0, sin(x)/(1 + cos x) should just go to 0/(1 + 1), or just 0.
lim x→ 0
[
1 − cos x x
]
= lim x→ 0
[
sin(x) x
sin(x) 1 + cos x
]
= lim x→ 0
[
sin(x) x
]
=
· lim x→ 0
[
sin(x) 1 + cos x
]
= = 1 · 0 = 0
So we’ve calculated both of the limits! We know:
lim h→ 0
sin h h
= 1 and lim h→ 0
1 − cos h h
So with these limits both calculated, we can return to our original argument, and finally prove that the derivative of sine is cosine! When we left off, we had:
sin′(x) = cos(x) · lim h→ 0
[
sin(h) h
]
− sin(x) · lim h→ 0
[
1 − cos(h) h
]
and if I plug in what we just found out about these limits...
= cos(x) · lim h→ 0
[
sin(h) h
]
=
− sin(x) · lim h→ 0
[
1 − cos(h) h
]
= = cos(x)· 1 − sin(x)· 0 = cos(x)
Problem
Prove that cos′(x) = − sin(x). There are a couple ways you could do this. One way would be to follow the method we used here—to plug cos(x) into the FDQ, do the trig addition identity, rearrange, and figure out the limits. You can probably get it just in terms of the two limits we evaluated here, so no need to do anything new to figure out what they are. Another way would be to first rewrite cosine using a trig identity, and then take its derivative. Try it both ways. Write them up nicely and turn it in.
