Proof (derivatives of sin/cos/tan)

[QuarkCharmer]

Homework Statement

Prove that:
$$\frac{d}{dx}sin(x) = cos(x)$$
$$\frac{d}{dx}tan(x) = sec^2(x)$$

Homework Equations

Difference Quotient, Trig Identities

The Attempt at a Solution

Not sure how to go about the rest of this?

I haven't attempted the tangent yet, but I will post it here when I do rather than 2 threads.

Edit: Latex for this one a bit over my head, I am uploading a picture...

 

[Mark44]

Homework Statement

Prove that:
$$\frac{d}{dx}sin(x) = cos(x)$$
$$\frac{d}{dx}tan(x) = sec^2(x)$$

Homework Equations

Difference Quotient, Trig Identities

The Attempt at a Solution

Edit: Latex for this one a bit over my head, I am uploading a picture...

As I recall, there are a couple of limits that come into play when you find the derivatives of these trig functions using the definition of the derivative.

$$\lim_{h \to 0} \frac{sin(h)}{h}=1$$
$$\lim_{h \to 0} \frac{tan(h)}{h}=1$$

 

[QuarkCharmer]

Yeah, L'Hopital's rule. But you have to know that d/dx sin(x) = cos(x), and that's what I'm trying to prove lol.

Either way, I still have something like sin(x) over the delta-x, so I am not sure that applies?

I'm just having a huge headache with these two problems and now I'm doubting the whole thing!

 

[Mark44]

Yeah, L'Hopital's rule. But you have to know that d/dx sin(x) = cos(x), and that's what I'm trying to prove lol.

No, these limits can be shown without using L'Hopital's Rule.

Either way, I still have something like sin(x) over the delta-x, so I am not sure that applies?

I'm just having a huge headache with these two problems and now I'm doubting the whole thing!

 

[QuarkCharmer]

Ugh, I think I see an error that I can't believe I made. Working on no sleep here. I'll give it another shot now.

 

[Mark44]

The other limit that is used is
$$\lim_{h \to 0} \frac{1 - cos(h)}{h}=0$$

 

[QuarkCharmer]

I'm not seeing how sin(h)/h = 1 as h tends to 0. Wouldn't that mean that sin(h) is approaching 1, but it's divided by something approaching zero? I know you can divide this way with limits because it's not 0, it's tending towards 0, but I'm still not seeing it?

 

[fzero]

I'm not seeing how sin(h)/h = 1 as h tends to 0. Wouldn't that mean that sin(h) is approaching 1, but it's divided by something approaching zero? I know you can divide this way with limits because it's not 0, it's tending towards 0, but I'm still not seeing it?

sin(h) approaches h as h tends to 0. It might be helpful graph y = sin x and y =x together near x=0.

 

[Mark44]

Here's why lim sinh/h = 1, as h --> 0.

One proof uses the unit circle. Extend a ray from the origin out to and past a point on the circle. The ray should be at a small angle h (in radians) to the x-axis.

The ray intersects the circle at (cos h, sin h). At this point, drop a vertical line down to the x-axis. At (1, 0), draw a vertical line segment up to the ray. That point of intersection is (1, tan h).

The ray defines a small right triangle, a circle sector, and a large right triangle. Looking at the areas, we see that the area of the small triangle <= area of the sector <= area of large triangle.

That is, (1/2) cos(h) sin(h) <= (1/2) h <= (1/2) tan(h), where (1/2) h is the area of the sector.

Equivalently, cos(h) sin(h) <= h <= tan(h), and this is true for h "small" and close to zero.

Divide by sin(h) to get:

cos(h) <= h/sin(h) <= 1/cos(h)

Use the squeeze theorem. cos(h) --> 1 and 1/cos(h) --> 1 as h --> 0, hence lim h/sin(h) = 1, as h --> 0.

 

[QuarkCharmer]

Not sure why I wrote sin(h) = 1 as h tends to 0 up there. In my head I was thinking 0, which still leads to a ratio that is approaching 1, but not one. I didn't understand how it could be approaching 1, but the tutoring center here helped me out with that same explanation. Thanks!

I'll try my hand at the same problem for tangent now.

Edit: That is a good squeeze theorem explanation above. I didn't see that post when I submitted this first, but that one is great. Many thanks.

 

[QuarkCharmer]

Alright, I think I have this down now. Would you mind looking it over?

 

[Mark44]

Your scanned image is hard to read. The bottom half is fairly clear, but the top half is pretty faint.

 

[QuarkCharmer]

Your scanned image is hard to read. The bottom half is fairly clear, but the top half is pretty faint.

Okay, I'll give this a shot:
Pretend there is a limit as h tends to 0 before all this:

$$\frac{d}{dx} tan(x) = sec^2(x)$$

$$\frac{tan(x+h)-tan(x)}{h} =$$

Here I just gave a common den. and used the tan ident.

$$\frac{\frac{tan(x)+tan(h)}{1-tan(x)tan(h)} - \frac{tan(x)(1-tan(x)tan(h))}{1-tan(x)tan(h)}}{h}$$

$$\frac{\frac{tan(h)+tan^2(x)tan(h)}{1-tan(x)tan(h)}}{h}$$

$$\frac{tan(h)(1+tan^2(x))(1)}{h(1-tan(x)tan(h))}$$

$$\frac{tan(h)}{h} \frac{sec^2(x)}{1-tan(x)tan(h)}$$

$$\frac{sec^2(x)}{1-0}$$

$$sec^2(x) = sec^2(x)$$

 

[Mark44]

Okay, I'll give this a shot:
Pretend there is a limit as h tends to 0 before all this:

I will also pretend that there is an equals between all of these equal expressions.

$$\frac{d}{dx} tan(x) = sec^2(x)$$

$$\frac{tan(x+h)-tan(x)}{h} =$$

Here I just gave a common den. and used the tan ident.

$$\frac{\frac{tan(x)+tan(h)}{1-tan(x)tan(h)} - \frac{tan(x)(1-tan(x)tan(h))}{1-tan(x)tan(h)}}{h}$$

$$\frac{\frac{tan(h)+tan^2(x)tan(h)}{1-tan(x)tan(h)}}{h}$$

$$\frac{tan(h)(1+tan^2(x))(1)}{h(1-tan(x)tan(h))}$$

$$\frac{tan(h)}{h} \frac{sec^2(x)}{1-tan(x)tan(h)}$$

$$\frac{sec^2(x)}{1-0}$$

$$sec^2(x) = sec^2(x)$$

Somewhere near the end you "passed to the limit" as they say, or took the limit.

$$= \lim_{h \to 0}\frac{tan(h)}{h} \frac{sec^2(x)}{1-tan(x)tan(h)}$$

$$= \lim_{h \to 0}\frac{tan(h)}{h} \lim_{h \to 0}\frac{sec^2(x)}{1-tan(x)tan(h)}$$

$$= 1 \cdot sec^2(x)$$

In the line above the last one, lim tan(h)/h = 1, and lim (sec^2(x))/(1 - tan(x)tan(h)) = sec^2(x), as h --> 0.

 

[QuarkCharmer]

Yeah, it wasn't really a "homework" assignment, nobody will read it. I was mostly curious myself more than anything. I really appreciate all of the help. Maybe one day someone will ask something that I can help with and I can repay.