# Basic calculus 1 question: What is the defining formula of sinΘ?

1. Mar 2, 2010

1. The problem statement, all variables and given/known data

Q) Derive the dereitvative of sinΘ using the definition of differentiation.

I am in Calculus 3, and I used to know how to work this problem very well! :) I just don't remember lol - I just need a little help I guess! :)

2. Relevant equations

lim [f(x+h) - f(x))]/h
x→0

3. The attempt at a solution

f(x) = sinΘ

f(x+h) = (sinΘ + h)

Now I am STUCK!! lol

2. Mar 2, 2010

### rock.freak667

well now put that into the formula and check the numerator.

You will have sin(θ+h)-sinθ. Do you know your sum to product formulas? These will help greatly here or you can expand sin(θ+h) as well.

3. Mar 2, 2010

### Staff: Mentor

Might as well use x instead of theta, since x is easier to type.

f(x + h) = sin(x + h)
Use the identity for the sine of the sum of two angles.

You will also need a limit:
$$\lim_{h \to 0} \frac{sin(h)}{h}$$
Do you remember the value of this limit?

4. Mar 2, 2010

so, I write:

lim [sin(x+h) - sin(x))]/h
x→0

So, I get:

lim [sin(x) cos(h) + cos(x)sin(h) - sin(x))]/h
x→0

But now I am stuck again lol - I understand that if I use $$\lim_{h \to 0} \frac{sin(h)}{h}$$

then my equation should become:

lim [sin(x) cos(h) + cos(x)sin(h) ]/h , right?
x→0

5. Mar 2, 2010

### Staff: Mentor

Your limit should be as h --> 0, not as x --> 0. Also, you are missing a term in the numerator. What happened to the -sin(x) that used to be there?

There's another limit that you'll need as well:
$$\lim_{h \to 0} \frac{cos(h) - 1}{h}$$

6. Mar 7, 2010